4.4 Trigonometrische Gleichungen

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{{Vald flik|[[4.4 Trigonometriska ekvationer|Teori]]}}
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{{Vald flik|[[4.4 Trigonometriska ekvationer|Theory]]}}
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{{Ej vald flik|[[4.4 Övningar|Övningar]]}}
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{{Ej vald flik|[[4.4 Övningar|Exercises]]}}
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Content:'''
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*Trigonometriska grundekvationer
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* The basic equations of trigonometry
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*Enklare trigonometriska ekvationer
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*Simple trigonometric equations
}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes: '''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned how to:
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*Lösa trigonometriska grundekvationer.
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* Solve the basic equations of trigonometry
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*Lösa trigonometriska ekvationer som kan återföras till ovanstående ekvationstyp.
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* Solve trigonometric equations that can be reduced to a basic equation.
}}
}}
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== Grundekvationer ==
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== Basic equations ==
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Trigonometriska ekvationer kan vara mycket komplicerade, men det finns också många typer av trigonometriska ekvationer som man kan lösa med ganska enkla metoder. Här skall vi börja med att titta på de mest grundläggande trigonometriska ekvationerna, av typerna <math>\sin x = a</math>, <math>\cos x = a</math> och <math>\tan x = a</math>.
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Trigonometric equations can be very complicated, but there are also many types of trigonometric equations which can be solved using relatively simple methods. Here, we shall start by looking at the most basic trigonometric equations, of the type <math>\sin x = a</math>, <math>\cos x = a</math> and <math>\tan x = a</math>.
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Dessa ekvationer har i regel oändligt många lösningar, såvida inte omständigheterna begränsar antalet möjliga lösningar (t.ex. att man söker en spetsig vinkel).
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These equations usually have an infinite number of solutions, unless the circumstances limit the number of possible solutions (for example, if one is looking for an acute angle).
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
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Lös ekvationen <math>\,\sin x = \frac{1}{2}</math>.
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Solve the equation <math>\,\sin x = \frac{1}{2}</math>.
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Vår uppgift är att bestämma alla vinklar som gör att sinus av vinkeln blir <math>\tfrac{1}{2}</math>. Vi tar hjälp av enhetscirkeln. Notera att vinkeln här kallas <math>x</math>.
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Our task is to determine all the angles that have a sine with the value <math>\tfrac{1}{2}</math>. The unit circle helps us in this . Note that here the angle is designated as <math>x</math>.
<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/6 resp. 5π/6}}</center>
<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/6 resp. 5π/6}}</center>
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I figuren har vi angivit de två riktningar som ger punkter med ''y''-koordinat <math>\tfrac{1}{2}</math> i enhetscirkeln, dvs. vinklar med sinusvärdet <math>\tfrac{1}{2}</math>. Den första är standardvinkeln <math>30^\circ = \pi / 6</math> och av symmetriskäl bildar den andra vinkeln <math>30^\circ</math> mot den negativa ''x''-axeln, vilket gör att den vinkeln är <math>180^\circ – 30^\circ = 150^\circ</math> eller i radianer <math>\pi – \pi / 6 = 5\pi / 6</math>. Detta är de enda lösningar till ekvationen <math>\sin x = \tfrac{1}{2}</math> mellan <math>0</math> och <math>2\pi</math>.
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In the figure, we have shown the two directions that give us points which have a ''y''- coordinate <math>\tfrac{1}{2}</math> on the unit circle, i.e. angles with a sine value <math>\tfrac{1}{2}</math>. The first is the standard angle <math>30^\circ = \pi / 6</math> and by symmetry the other angle makes <math>30^\circ</math> with the negative ''x''-axis, This means that the angle is <math>180^\circ – 30^\circ = 150^\circ</math> or in radians <math>\pi – \pi / 6 = 5\pi / 6</math>. These are the only solutions to the equation <math>\sin x = \tfrac{1}{2}</math> between <math>0</math> and <math>2\pi</math>.
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Vi kan dock lägga till ett godtyckligt antal varv till dessa två vinklar och fortfarande få samma sinusvärde. Alla vinklar med sinusvärde <math>\tfrac{1}{2}</math> är alltså
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However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles with a value of the sine <math>\tfrac{1}{2}</math> are
{{Fristående formel||<math>\begin{cases}
{{Fristående formel||<math>\begin{cases}
x &= \dfrac{\pi}{6} + 2n\pi\\
x &= \dfrac{\pi}{6} + 2n\pi\\
x &= \dfrac{5\pi}{6} + 2n\pi
x &= \dfrac{5\pi}{6} + 2n\pi
\end{cases}</math>}}
\end{cases}</math>}}
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där <math>n</math> är ett godtyckligt heltal. Detta kallas för den fullständiga lösningen till ekvationen.
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where <math>n</math> is an arbitrary integer. This is called the general solution to the equation.
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Lösningarna syns också i figuren nedan där grafen till <math>y = \sin x</math> skär linjen <math>y=\tfrac{1}{2}</math>.
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The solutions also seen in the figure below where the graph of <math>y = \sin x</math> intersect the line <math>y=\tfrac{1}{2}</math>.
<center>{{:4.4 - Figur - Kurvorna y = sin x och y = ½}}</center>
<center>{{:4.4 - Figur - Kurvorna y = sin x och y = ½}}</center>
Zeile 53: Zeile 53:
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
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Lös ekvationen <math>\,\cos x = \frac{1}{2}</math>.
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Solve the equation <math>\,\cos x = \frac{1}{2}</math>.
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Vi tar återigen hjälp av enhetscirkeln.
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We once again study the unit circle.
<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. -π/3}}</center>
<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. -π/3}}</center>
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Vi vet att cosinus blir <math>\tfrac{1}{2}</math> för vinkeln <math>\pi/3</math>. Den enda andra riktning i enhetscirkeln som ger samma värde på cosinus har vinkeln <math>-\pi/3</math>. Lägger vi till ett helt antal varv till dessa vinklar får vi den fullständiga lösningen
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We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other direction in the unit circle, which produces the same value for the cosine is the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution
{{Fristående formel||<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}}
{{Fristående formel||<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}}
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där <math>n</math> är ett godtyckligt heltal.
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where <math>n</math> is an arbitrary integer.
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
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Lös ekvationen <math>\,\tan x = \sqrt{3}</math>.
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Solve the equation <math>\,\tan x = \sqrt{3}</math>.
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En lösning till ekvationen är standardvinkeln <math>x=\pi/3</math>.
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A solution to the equation is the standard angle <math>x=\pi/3</math>.
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Om vi betraktar enhetscirkeln så är tangens av en vinkel lika med riktningskoefficienten för den räta linje genom origo som bildar vinkeln <math>x</math> med den positiva ''x''-axeln.
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If we study the unit circle then we see that tangent of an angle is equal to the slope of the straight line through the origin making an angle <math>x</math> with the positive ''x''-axis .
<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. π+π/3}}</center>
<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. π+π/3}}</center>
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Därför ser vi att lösningarna till <math>\tan x = \sqrt{3}</math> upprepar sig varje halvt varv <math>\pi/3</math>, <math>\pi/3 +\pi</math>, <math>\pi/3+ \pi +\pi</math> osv. Den fullständiga lösningen kan vi därmed få fram genom att utgå från lösningen <math>\pi/3</math> och lägga till eller dra ifrån multiplar av <math>\pi</math>,
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Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution <math>\pi/3</math>, <math>\pi/3 +\pi</math>, <math>\pi/3+ \pi +\pi</math> and so on. The general solution can be obtained by using the solution <math>\pi/3</math> and adding or subtracting multiples of <math>\pi</math>,
{{Fristående formel||<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}
{{Fristående formel||<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}
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där <math>n</math> är ett godtyckligt heltal.
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where <math>n</math> s an arbitrary integer.
</div>
</div>
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== Några mer komplicerade ekvationer ==
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== Somewhat more complicated equations ==
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Trigonometriska ekvationer kan se ut på många olika sätt, och det är omöjligt att här ge en fullständig genomgång av alla tänkbara ekvationer. Men låt oss studera några exempel, där vi kan ha nytta av att vi kan lösa grundekvationerna.
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Trigonometric equations can vary in many ways, and it is impossible to give a full catalogue of all possible equations. But let us study some examples where we can use our knowledge of solving basic equations.
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Vissa trigonometriska ekvationer kan förenklas genom att de skrivs om med hjälp av trigonometriska samband. Detta kan t.ex. leda till en andragradsekvation, som i nedanstående exempel där man använder att <math>\cos 2x = 2 \cos^2\!x – 1</math>.
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Some trigonometric equations can be simplified by being rewritten with the help of trigonometric relationships. This, for example, could lead to a quadratic equations, as in the example below where one uses <math>\cos 2x = 2 \cos^2\!x – 1</math>.
<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
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Lös ekvationen <math>\,\cos 2x – 4\cos x + 3= 0</math>.
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Solve the equation <math>\,\cos 2x – 4\cos x + 3= 0</math>.
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Omskrivning med hjälp av formeln <math>\cos 2x = 2 \cos^2\!x – 1</math> ger
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Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math> giving
{{Fristående formel||<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{,}</math>}}
{{Fristående formel||<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{,}</math>}}
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vilket kan förenklas till ekvationen (efter division med 2)
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which can be simplified to the equation (after division by 2)
{{Fristående formel||<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}}
{{Fristående formel||<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}}
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Vänsterledet kan faktoriseras med kvadreringsregeln till
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The left-hand side can factorised by using the squaring rule to give
{{Fristående formel||<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}}
{{Fristående formel||<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}}
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Denna ekvation kan bara vara uppfylld om <math>\cos x = 1</math>. Grundekvationen <math>\cos x=1</math> kan vi lösa på det vanliga sättet och den fullständiga lösningen är
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This equation can only be satisfied if <math>\cos x = 1</math>. The basic equation <math>\cos x=1</math> can be solved in the normal way and the complete solution is
{{Fristående formel||<math>
{{Fristående formel||<math>
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x = 2n\pi \qquad (\,n \mbox{ godtyckligt heltal).}</math>}}
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x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
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Lös ekvationen <math>\,\frac{1}{2}\sin x + 1 – \cos^2 x = 0</math>.
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Solve the equation <math>\,\frac{1}{2}\sin x + 1 – \cos^2 x = 0</math>.
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Enligt den trigonometriska ettan är <math>\sin^2\!x + \cos^2\!x = 1</math>, dvs. <math>1 – \cos^2\!x = \sin^2\!x</math>.
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According to the Pythagorean identity <math>\sin^2\!x + \cos^2\!x = 1</math>, i.e.. <math>1 – \cos^2\!x = \sin^2\!x</math>.
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Ekvationen kan alltså skrivas
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The equation can be written as
{{Fristående formel||<math>\tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}</math>}}
{{Fristående formel||<math>\tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}</math>}}
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Genom att nu bryta ut en faktor <math>\sin x</math> får vi
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Factorising out <math>\sin x</math> one gets
{{Fristående formel||<math>
{{Fristående formel||<math>
\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}
\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}
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Från denna faktoriserade form av ekvationen ser vi att lösningarna antingen måste uppfylla <math>\sin x = 0</math> eller <math>\sin x = -\tfrac{1}{2}</math>, vilka är två vanliga grundekvationer på formen <math>\sin x = a</math> och kan lösas som i exempel 1. Lösningarna blir till slut
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From this factorised form of the equation, we see that the solutions either have to satisfy <math>\sin x = 0</math> or <math>\sin x = -\tfrac{1}{2}</math>, which are two basic equations of the type <math>\sin x = a</math> and can be solved as in Example 1. The solutions turn out to be
{{Fristående formel||<math>
{{Fristående formel||<math>
\begin{cases}
\begin{cases}
Zeile 142: Zeile 142:
<div class="exempel">
<div class="exempel">
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'''Exempel 6'''
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''' Example 6'''
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Lös ekvationen <math>\,\sin 2x =4 \cos x</math>.
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Solve the equation <math>\,\sin 2x =4 \cos x</math>.
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Genom omskrivning med formeln för dubbla vinkeln blir ekvationen
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By rewriting the equation using the formula for double-angles one gets
{{Fristående formel||<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}}
{{Fristående formel||<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}}
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Vi delar båda led med 2 och bryter ut en faktor <math>\cos x</math>, vilket ger
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We divide both sides with 2 and factorise out <math>\cos x</math>, which gives
{{Fristående formel||<math>\cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}</math>}}
{{Fristående formel||<math>\cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}</math>}}
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Eftersom produkten i vänsterledet bara kan bli noll genom att en faktor är noll, så kan ekvationen delas upp i grundekvationerna
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As the product of factors on the left-hand side can only be zero if one of the factors is zero, we have reduced the original equation into two basic equations
* <math>\cos x = 0</math>,
* <math>\cos x = 0</math>,
* <math>\sin x = 2</math>.
* <math>\sin x = 2</math>.
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Men <math>\sin x</math> kan aldrig bli större än 1, så ekvationen <math>\sin x = 2</math> saknar lösningar. Då återstår bara
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But <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That leaves just
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<math>\cos x = 0</math>, vilken med hjälp av enhetscirkeln ger den fullständiga lösningen <math>x = \pi / 2 + n \cdot \pi</math>.
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<math>\cos x = 0</math>, and using the unit circle gives the general solution <math>x = \pi / 2 + n \cdot \pi</math>.
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 7'''
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''' Example 7'''
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Lös ekvationen <math>\,4\sin^2\!x – 4\cos x = 1</math>.
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Solve the equation <math>\,4\sin^2\!x – 4\cos x = 1</math>.
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Med den trigonometriska ettan kan <math>\sin^2\!x</math> bytas ut mot <math>1 – \cos^2\!x</math>. Då får vi
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Using the Pythagorean identity one can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then we will have{{Fristående formel||<math>
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{{Fristående formel||<math>
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\begin{align*}
\begin{align*}
4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\
4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\
Zeile 176: Zeile 175:
\end{align*}</math>}}
\end{align*}</math>}}
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Detta är en andragradsekvation i <math>\cos x</math>, som har lösningarna
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This is a quadratic equation in <math>\cos x</math>, which has the solutions
{{Fristående formel||<math>
{{Fristående formel||<math>
\cos x = -\tfrac{3}{2} \quad\text{och}\quad
\cos x = -\tfrac{3}{2} \quad\text{och}\quad
\cos x = \tfrac{1}{2}\,\mbox{.}</math>}}
\cos x = \tfrac{1}{2}\,\mbox{.}</math>}}
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Eftersom värdet av <math>\cos x</math> ligger mellan <math>–1</math> och <math>1</math> kan ekvationen <math>\cos x=-\tfrac{3}{2}</math> inte ha några lösningar. Då återstår bara grundekvationen
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Since the value of <math>\cos x</math> is between <math>–1</math> and <math>1</math> the equation <math>\cos x=-\tfrac{3}{2}</math> has no solutions. That leaves only the basic equation
{{Fristående formel||<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}
{{Fristående formel||<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}
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som löses enligt exempel 2.
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that may be solved as in example 2.
</div>
</div>
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[[4.4 Övningar|Övningar]]
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[[4.4 Övningar|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
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'''Råd för inläsning'''
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'''Study advice'''
 +
 
 +
'''Basic and final tests'''
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'''Grund- och slutprov'''
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After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
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Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
 
 +
'''Remember:'''
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'''Tänk på att:'''
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It is good idea to learns the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.
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Det är bra om man lär sig de vanliga trigonometriska formlerna (identiteterna) och övar upp en viss vana på att förenkla och manipulera trigonometriska uttryck.
 
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Det är viktigt att man lär sig de grundläggande ekvationerna, av typen <math>\sin x = a</math>, <math>\cos x = a</math> eller <math>\tan x = a</math> (där <math>a</math> är ett reellt tal). Det är också viktigt att man vet att dessa ekvationer typiskt har oändligt många lösningar.
+
It is important to be familiar with the basic equations, such as <math>\sin x = a</math>, <math>\cos x = a</math> or <math>\tan x = a</math> (where <math>a</math> is a real number). It is also important to know that these equations typically have infinitely many solutions.
-
'''Lästips'''
+
'''Reviews'''
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för dig som vill fördjupa dig ytterligare eller behöver en längre förklaring vill vi tipsa om:
+
For those of you who want to deepen your studies or need more detailed explanations consider the following references
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[http://www.theducation.se/kurser/umaprep/4_trigonometri/44_trig_ekvationer/index.asp Läs mer om trigonometriska ekvationer i Theducations gymnasielexikon ]
+
[http://www.theducation.se/kurser/umaprep/4_trigonometri/44_trig_ekvationer/index.asp Learn more about trigonometric equations in Theducations gymnasielexikon ]
-
[http://www.theducation.se/kurser/umaprep/4_trigonometri/44_trig_ekvationer/445_typ_asinx/index.asp Träna på trigonometriska räkneexempel i Theducations gymnasielexikon]
+
[http://www.theducation.se/kurser/umaprep/4_trigonometri/44_trig_ekvationer/445_typ_asinx/index.asp Practise trigonometric calculations in Theducations gymnasielexikon ]
'''Länktips'''
'''Länktips'''
-
[http://www.ies.co.jp/math/java/trig/ABCsinX/ABCsinX.html Experimentera med grafen y=a sin b(x-c)]
+
[http://www.ies.co.jp/math/java/trig/ABCsinX/ABCsinX.html Experiment with the graph y = a sin b (x-c) ]
-
[http://www.theducation.se/kurser/experiment/gyma/applets/ex45_derivatasinus/Ex45Applet.html Experimentera med derivatan av sin x]
+
[http://www.theducation.se/kurser/experiment/gyma/applets/ex45_derivatasinus/Ex45Applet.html Experiment with the derivative of sin x]
</div>
</div>

Version vom 18:28, 16. Jul. 2008

 

Vorlage:Vald flik Vorlage:Ej vald flik

 

Content:

  • The basic equations of trigonometry
  • Simple trigonometric equations

Learning outcomes:

After this section, you will have learned how to:

  • Solve the basic equations of trigonometry
  • Solve trigonometric equations that can be reduced to a basic equation.

Basic equations

Trigonometric equations can be very complicated, but there are also many types of trigonometric equations which can be solved using relatively simple methods. Here, we shall start by looking at the most basic trigonometric equations, of the type \displaystyle \sin x = a, \displaystyle \cos x = a and \displaystyle \tan x = a.

These equations usually have an infinite number of solutions, unless the circumstances limit the number of possible solutions (for example, if one is looking for an acute angle).

Example 1

Solve the equation \displaystyle \,\sin x = \frac{1}{2}.


Our task is to determine all the angles that have a sine with the value \displaystyle \tfrac{1}{2}. The unit circle helps us in this . Note that here the angle is designated as \displaystyle x.

4.4 - Figur - Två enhetscirklar med vinklar π/6 resp. 5π/6

In the figure, we have shown the two directions that give us points which have a y- coordinate \displaystyle \tfrac{1}{2} on the unit circle, i.e. angles with a sine value \displaystyle \tfrac{1}{2}. The first is the standard angle \displaystyle 30^\circ = \pi / 6 and by symmetry the other angle makes \displaystyle 30^\circ with the negative x-axis, This means that the angle is \displaystyle 180^\circ – 30^\circ = 150^\circ or in radians \displaystyle \pi – \pi / 6 = 5\pi / 6. These are the only solutions to the equation \displaystyle \sin x = \tfrac{1}{2} between \displaystyle 0 and \displaystyle 2\pi.

However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles with a value of the sine \displaystyle \tfrac{1}{2} are Vorlage:Fristående formel where \displaystyle n is an arbitrary integer. This is called the general solution to the equation.

The solutions also seen in the figure below where the graph of \displaystyle y = \sin x intersect the line \displaystyle y=\tfrac{1}{2}.

4.4 - Figur - Kurvorna y = sin x och y = ½

Example 2

Solve the equation \displaystyle \,\cos x = \frac{1}{2}.


We once again study the unit circle.

4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. -π/3

We know that cosine is \displaystyle \tfrac{1}{2} for the angle \displaystyle \pi/3. The only other direction in the unit circle, which produces the same value for the cosine is the angle \displaystyle -\pi/3. Adding an integral number of revolutions to these angles we get the general solution

Vorlage:Fristående formel

where \displaystyle n is an arbitrary integer.

Example 3

Solve the equation \displaystyle \,\tan x = \sqrt{3}.


A solution to the equation is the standard angle \displaystyle x=\pi/3.

If we study the unit circle then we see that tangent of an angle is equal to the slope of the straight line through the origin making an angle \displaystyle x with the positive x-axis .

4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. π+π/3

Therefore, we see that the solutions to \displaystyle \tan x = \sqrt{3} repeat themselves every half revolution \displaystyle \pi/3, \displaystyle \pi/3 +\pi, \displaystyle \pi/3+ \pi +\pi and so on. The general solution can be obtained by using the solution \displaystyle \pi/3 and adding or subtracting multiples of \displaystyle \pi, Vorlage:Fristående formel

where \displaystyle n s an arbitrary integer.


Somewhat more complicated equations

Trigonometric equations can vary in many ways, and it is impossible to give a full catalogue of all possible equations. But let us study some examples where we can use our knowledge of solving basic equations.

Some trigonometric equations can be simplified by being rewritten with the help of trigonometric relationships. This, for example, could lead to a quadratic equations, as in the example below where one uses \displaystyle \cos 2x = 2 \cos^2\!x – 1.

Example 4

Solve the equation \displaystyle \,\cos 2x – 4\cos x + 3= 0.


Rewrite by using the formula \displaystyle \cos 2x = 2 \cos^2\!x – 1 giving Vorlage:Fristående formel

which can be simplified to the equation (after division by 2)

Vorlage:Fristående formel

The left-hand side can factorised by using the squaring rule to give

Vorlage:Fristående formel

This equation can only be satisfied if \displaystyle \cos x = 1. The basic equation \displaystyle \cos x=1 can be solved in the normal way and the complete solution is

Vorlage:Fristående formel

Example 5

Solve the equation \displaystyle \,\frac{1}{2}\sin x + 1 – \cos^2 x = 0.


According to the Pythagorean identity \displaystyle \sin^2\!x + \cos^2\!x = 1, i.e.. \displaystyle 1 – \cos^2\!x = \sin^2\!x. The equation can be written as Vorlage:Fristående formel

Factorising out \displaystyle \sin x one gets Vorlage:Fristående formel

From this factorised form of the equation, we see that the solutions either have to satisfy \displaystyle \sin x = 0 or \displaystyle \sin x = -\tfrac{1}{2}, which are two basic equations of the type \displaystyle \sin x = a and can be solved as in Example 1. The solutions turn out to be Vorlage:Fristående formel

Example 6

Solve the equation \displaystyle \,\sin 2x =4 \cos x.


By rewriting the equation using the formula for double-angles one gets Vorlage:Fristående formel

We divide both sides with 2 and factorise out \displaystyle \cos x, which gives Vorlage:Fristående formel

As the product of factors on the left-hand side can only be zero if one of the factors is zero, we have reduced the original equation into two basic equations

  • \displaystyle \cos x = 0,
  • \displaystyle \sin x = 2.

But \displaystyle \sin x can never be greater than 1, so the equation \displaystyle \sin x = 2 has no solutions. That leaves just \displaystyle \cos x = 0, and using the unit circle gives the general solution \displaystyle x = \pi / 2 + n \cdot \pi.

Example 7

Solve the equation \displaystyle \,4\sin^2\!x – 4\cos x = 1.


Using the Pythagorean identity one can replace \displaystyle \sin^2\!x by \displaystyle 1 – \cos^2\!x. Then we will haveVorlage:Fristående formel

This is a quadratic equation in \displaystyle \cos x, which has the solutions Vorlage:Fristående formel

Since the value of \displaystyle \cos x is between \displaystyle –1 and \displaystyle 1 the equation \displaystyle \cos x=-\tfrac{3}{2} has no solutions. That leaves only the basic equation Vorlage:Fristående formel

that may be solved as in example 2.


Exercises

Study advice

Basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Remember:

It is good idea to learns the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.


It is important to be familiar with the basic equations, such as \displaystyle \sin x = a, \displaystyle \cos x = a or \displaystyle \tan x = a (where \displaystyle a is a real number). It is also important to know that these equations typically have infinitely many solutions.


Reviews

For those of you who want to deepen your studies or need more detailed explanations consider the following references

Learn more about trigonometric equations in Theducations gymnasielexikon

Practise trigonometric calculations in Theducations gymnasielexikon


Länktips

Experiment with the graph y = a sin b (x-c)

Experiment with the derivative of sin x

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