4.4 Trigonometric equations

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m
Current revision (16:46, 6 January 2009) (edit) (undo)
m ("solution to" becomes "solution of"; some other minor changes)
 
(24 intermediate revisions not shown.)
Line 2: Line 2:
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" |  
| style="border-bottom:1px solid #797979" width="5px" |  
-
{{Vald flik|[[4.4 Trigonometriska ekvationer|Theory]]}}
+
{{Selected tab|[[4.4 Trigonometric equations|Theory]]}}
-
{{Ej vald flik|[[4.4 Övningar|Exercises]]}}
+
{{Not selected tab|[[4.4 Exercises|Exercises]]}}
| style="border-bottom:1px solid #797979" width="100%"|  
| style="border-bottom:1px solid #797979" width="100%"|  
|}
|}
{{Info|
{{Info|
-
'''Content:'''
+
'''Contents:'''
-
* The basic equations of trigonometry
+
* Simple trigonometric equations
-
*Simple trigonometric equations
+
}}
}}
Line 17: Line 16:
After this section, you will have learned how to:
After this section, you will have learned how to:
-
* Solve the basic equations of trigonometry
+
* Solve the basic equations of trigonometry.
-
* Solve trigonometric equations that can be reduced to a basic equation.
+
* Solve trigonometric equations that can be reduced to basic equations.
}}
}}
== Basic equations ==
== 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 <math>\sin x = a</math>, <math>\cos x = a</math> and <math>\tan x = a</math>.
+
Trigonometric equations can be very complicated, but there are also many types 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>.
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).
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).
Line 33: Line 32:
-
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>.
+
Our task is to determine all the angles that have a sine equal to <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 - Figure - Two unit circles with angles π/6 and 5π/6, respectively}}</center>
-
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>.
+
The figure illustrates the two points on the circle which have ''y''-coordinate <math>\tfrac{1}{2}</math>, i.e. the two points whose corresponding angles have 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. Therefore the other 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>.
-
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
+
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 <math>x</math> for which <math>\sin x = \tfrac{1}{2}</math> are
-
{{Fristående formel||<math>\begin{cases}
+
{{Displayed math||<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
Line 46: Line 45:
where <math>n</math> is an arbitrary integer. This is called the general solution to the equation.
where <math>n</math> 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 <math>y = \sin x</math> intersect the line <math>y=\tfrac{1}{2}</math>.
+
The solutions can also be obtained from the figure below, by looking at where the graph of <math>y = \sin x</math> intersects the line <math>y=\tfrac{1}{2}</math>.
-
<center>{{:4.4 - Figur - Kurvorna y = sin x och y = ½}}</center>
+
<center>{{:4.4 - Figure - The curves y = sin x and y = ½}}</center>
</div>
</div>
Line 60: Line 59:
We once again study the unit circle.
We once again study the unit circle.
-
<center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. -π/3}}</center>
+
<center>{{:4.4 - Figure - Two unit circles with angles π/3 and -π/3, respectively}}</center>
-
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
+
We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other point on the unit circle which has ''x''-coordinate <math>\frac{1}{2}</math> corresponds to 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>}}
+
{{Displayed math||<math>x = \pm \pi/3 + n \times 2\pi\,\mbox{,}</math>}}
where <math>n</math> is an arbitrary integer.
where <math>n</math> is an arbitrary integer.
Line 75: Line 74:
-
A solution to the equation is the standard angle <math>x=\pi/3</math>.
+
One solution to the equation is the standard angle <math>x=\pi/3</math>.
-
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 .
+
If we study the unit circle then we see that the tangent of an angle is equal to the slope of the straight line through the origin which makes 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 - Figure - Two unit circles with angles π/3 and π+π/3, respectively}}</center>
-
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>,
+
Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution, and so the general solution can be obtained from the solution <math>\pi/3</math> by adding or subtracting multiples of <math>\pi</math>:
-
{{Fristående formel||<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}
+
{{Displayed math||<math>x = \pi/3 + n \, \pi\,\mbox{,}</math>}}
-
where <math>n</math> s an arbitrary integer.
+
where <math>n</math> is an arbitrary integer.
</div>
</div>
Line 92: Line 91:
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.
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 <math>\cos 2x = 2 \cos^2\!x – 1</math>.
+
Some trigonometric equations can be simplified by rewriting them with the help of the trigonometric relationships. This, for example, could lead to a quadratic equation, as in the example below.
<div class="exempel">
<div class="exempel">
Line 100: Line 99:
-
Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math> giving
+
Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math>, so that
-
{{Fristående formel||<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{,}</math>}}
+
{{Displayed math||<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{.}</math>}}
-
which can be simplified to the equation (after division by 2)
+
Simplifying, we see that
-
{{Fristående formel||<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}}
+
{{Displayed math||<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}}
The left-hand side can factorised by using the squaring rule to give
The left-hand side can factorised by using the squaring rule to give
-
{{Fristående formel||<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}}
+
{{Displayed math||<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}}
-
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
+
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 thus the complete solution is
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}</math>}}
x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}</math>}}
</div>
</div>
Line 123: Line 122:
-
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>.
+
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>. Thus the equation can be written as
-
The equation can be written as
+
{{Displayed math||<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>}}
+
-
Factorising out <math>\sin x</math> one gets
+
Factorising out <math>\sin x</math> we get
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
-
\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}
+
\sin x\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}
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
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>
+
{{Displayed math||<math>
\begin{cases}
\begin{cases}
x &= n\pi\\
x &= n\pi\\
Line 147: Line 145:
-
By rewriting the equation using the formula for double-angles one gets
+
By rewriting the equation using the formula for double-angles we get
-
{{Fristående formel||<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}}
+
{{Displayed math||<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}}
-
We divide both sides with 2 and factorise out <math>\cos x</math>, which gives
+
Dividing both sides by 2 and factorising out <math>\cos x</math>, gives
-
{{Fristående formel||<math>\cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}</math>}}
+
{{Displayed math||<math>\cos x\,( \sin x – 2) = 0\,\mbox{.}</math>}}
-
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
+
The left-hand side can only be zero if one of the factors is zero, and 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>.
-
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
+
However, <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That just leaves <math>\cos x = 0</math>, and using the unit circle we see that the general solution is
-
<math>\cos x = 0</math>, and using the unit circle gives the general solution <math>x = \pi / 2 + n \cdot \pi</math>.
+
{{Displayed math||<math>x = \frac{\pi}{2} + n\pi\mbox{,}</math>}}
 +
for <math>n</math> an arbitrary integer.
 +
 
</div>
</div>
Line 167: Line 167:
-
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>
+
Using the Pythagorean identity we can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then{{Displayed math||<math>
\begin{align*}
\begin{align*}
-
4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\
+
4 (1 – \cos^2\!x)\ –\ 4 \cos x &= 1\,\\
-
4 – 4 \cos^2\!x – 4 \cos x &= 1\,\mbox{,}\\
+
\Rightarrow 4\ –\ 4 \cos^2\!x\ –\ 4 \cos x &= 1\,\\
-
–4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\mbox{,}\\
+
\Rightarrow –4\cos^2\!x\ –\ 4 \cos x\ +\ 4\ –\ 1 &= 0\,\\
-
\cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
+
\Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
\end{align*}</math>}}
\end{align*}</math>}}
This is a quadratic equation in <math>\cos x</math>, which has the solutions
This is a quadratic equation in <math>\cos x</math>, which has the solutions
-
{{Fristående formel||<math>
+
{{Displayed math||<math>
-
\cos x = -\tfrac{3}{2} \quad\text{och}\quad
+
\cos x = -\tfrac{3}{2} \quad\text{and}\quad
\cos x = \tfrac{1}{2}\,\mbox{.}</math>}}
\cos x = \tfrac{1}{2}\,\mbox{.}</math>}}
-
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
+
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>}}
+
{{Displayed math||<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}
-
that may be solved as in example 2.
+
which may be solved as in Example 2.
</div>
</div>
-
[[4.4 Övningar|Exercises]]
+
[[4.4 Exercises|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
Line 197: Line 197:
-
'''Remember:'''
+
'''Keep in mind that...'''
-
It is good idea to learns the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.
+
It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.
Line 205: Line 205:
-
'''Reviews'''
+
'''Useful web sites'''
-
 
+
-
For those of you who want to deepen your studies or need more detailed explanations consider the following references
+
-
 
+
-
[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 Practise trigonometric calculations in Theducations gymnasielexikon ]
+
-
 
+
-
 
+
-
'''Länktips'''
+
[http://www.ies.co.jp/math/java/trig/ABCsinX/ABCsinX.html Experiment with the graph 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 Experiment with the derivative of sin x]
 
</div>
</div>

Current revision

       Theory          Exercises      

Contents:

  • 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 basic equations.

Basic equations

Trigonometric equations can be very complicated, but there are also many types 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 equal to \displaystyle \tfrac{1}{2}. The unit circle helps us in this. Note that here the angle is designated as \displaystyle x.

[Image]

The figure illustrates the two points on the circle which have y-coordinate \displaystyle \tfrac{1}{2}, i.e. the two points whose corresponding angles have 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. Therefore the other 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 \displaystyle x for which \displaystyle \sin x = \tfrac{1}{2} are

\displaystyle \begin{cases} 
   x &= \dfrac{\pi}{6} + 2n\pi\\
   x &= \dfrac{5\pi}{6} + 2n\pi
 \end{cases}

where \displaystyle n is an arbitrary integer. This is called the general solution to the equation.

The solutions can also be obtained from the figure below, by looking at where the graph of \displaystyle y = \sin x intersects the line \displaystyle y=\tfrac{1}{2}.

[Image]

Example 2

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


We once again study the unit circle.

[Image]

We know that cosine is \displaystyle \tfrac{1}{2} for the angle \displaystyle \pi/3. The only other point on the unit circle which has x-coordinate \displaystyle \frac{1}{2} corresponds to the angle \displaystyle -\pi/3. Adding an integral number of revolutions to these angles we get the general solution

\displaystyle x = \pm \pi/3 + n \times 2\pi\,\mbox{,}

where \displaystyle n is an arbitrary integer.

Example 3

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


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

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

[Image]

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

\displaystyle x = \pi/3 + n \, \pi\,\mbox{,}

where \displaystyle n is 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 rewriting them with the help of the trigonometric relationships. This, for example, could lead to a quadratic equation, as in the example below.

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, so that

\displaystyle (2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{.}

Simplifying, we see that

\displaystyle \cos^2\!x - 2 \cos x +1 =0\,\mbox{.}

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

\displaystyle (\cos x-1)^2 = 0\,\mbox{.}

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 thus the complete solution is

\displaystyle 
 x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}

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. Thus the equation can be written as

\displaystyle \tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}

Factorising out \displaystyle \sin x we get

\displaystyle 
 \sin x\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}

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

\displaystyle 
 \begin{cases}
   x &= n\pi\\
   x &= -\pi/6+2n\pi\\
   x &= 7\pi/6+2n\pi
 \end{cases}
 \qquad (\,n\ \text{ arbitrary integer})\mbox{.}

Example 6

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


By rewriting the equation using the formula for double-angles we get

\displaystyle 2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}

Dividing both sides by 2 and factorising out \displaystyle \cos x, gives

\displaystyle \cos x\,( \sin x – 2) = 0\,\mbox{.}

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

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

However, \displaystyle \sin x can never be greater than 1, so the equation \displaystyle \sin x = 2 has no solutions. That just leaves \displaystyle \cos x = 0, and using the unit circle we see that the general solution is

\displaystyle x = \frac{\pi}{2} + n\pi\mbox{,}

for \displaystyle n an arbitrary integer.

Example 7

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


Using the Pythagorean identity we can replace \displaystyle \sin^2\!x by \displaystyle 1 – \cos^2\!x. Then

\displaystyle 
 \begin{align*}
   4 (1 – \cos^2\!x)\ –\ 4 \cos x &= 1\,\\
   \Rightarrow 4\ –\ 4 \cos^2\!x\ –\ 4 \cos x &= 1\,\\
   \Rightarrow –4\cos^2\!x\ –\ 4 \cos x\ +\ 4\ –\ 1 &= 0\,\\
   \Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
 \end{align*}

This is a quadratic equation in \displaystyle \cos x, which has the solutions

\displaystyle 
 \cos x = -\tfrac{3}{2} \quad\text{and}\quad
 \cos x = \tfrac{1}{2}\,\mbox{.}

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

\displaystyle \cos x = \tfrac{1}{2}\,\mbox{,}

which 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.


Keep in mind that...

It is a good idea to learn 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.


Useful web sites

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

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/4.4_Trigonometric_equations"