4.4 Trigonometrische Gleichungen

Beispiel 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.

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

   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 in the figure below where the graph of \displaystyle y = \sin x intersects the line \displaystyle y=\tfrac{1}{2}.

Beispiel 2

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

We once again study the unit circle.

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

where \displaystyle n is an arbitrary integer.

Beispiel 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 .

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,

where \displaystyle n s an arbitrary integer.

Beispiel 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

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

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

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

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

Beispiel 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

Factorising out \displaystyle \sin x one gets

 \sin x\,\cdot\,\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

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

Beispiel 6

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

By rewriting the equation using the formula for double-angles one gets

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

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.

Beispiel 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 have

 \begin{align*}
   4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\
   4 – 4 \cos^2\!x – 4 \cos x &= 1\,\mbox{,}\\
   –4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\mbox{,}\\
   \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

 \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

that may be solved as in example 2.