5.3 | Solving Trigonometric Equations


Solving trigonometric equations in general is difficult. In this section we present techniques to solve trigonometric equations of particular forms. Some of these techniques are similar to those used to solve polynomial equations. The first method actually involves solving polynomial equations.

Although the three methods presented below solve some trigonometric equations, they do not solve all trigonometric equation.

Rewrite as a Polynomial Equation: If the given trigonometric equation is given such that replacing the trigonometric function with the variable \(u\) (or some variable that doesn’t appear in the given equation) results in a polynomial equation, then solve the corresponding polynomial equation for \(u\). Then substitute back the trigonometric function and either use the unit circle or the appropriate inverse trigonometric function to solve for the independent variable.

Solve
$$
\cos (3x-4) + 1 = -\cos (3x-4)
$$


Let \(u = \cos(3x-4)\), so that
$$
u + 1 = -u
$$
or
$$
u = \frac{-1}{2}.
$$
So,
$$
\cos (3x-4) = \frac{-1}{2}
$$
From the unit circle, we have that
$$
3x-4 = \pm\frac{5\pi}{6}+2n\pi,
$$
or
$$
x = \pm\frac{5\pi}{18}+\frac{4}{3}+\frac{2}{3}n\pi.
$$
Solve
$$
\frac{7}{4}\tan^2(4x-\pi)-\frac{5}{2}= -\frac{9}{4}
$$


Let \(u = \tan(4x-\pi)\). Then
$$
\frac{7}{4}u^2-\frac{5}{2}= -\frac{9}{4}
$$
or
$$
u = \pm \frac{1}{\sqrt{7}}
$$
So,
$$
\tan(4x-\pi) = \pm \frac{1}{\sqrt{7}}.
$$

The unit circle won’t help here, so we make use of inverse trigonometric functions. So,
$$
4x-\pi = \tan^{-1}(1/\sqrt{7})
$$
or
$$
4x-\pi = \tan^{-1}(-1/\sqrt{7})
$$
That is,
$$
x = \frac{\tan^{-1}(1/\sqrt{7})+\pi}{4} + \frac{1}{4}n\pi
$$
or
$$
x = \frac{\tan^{-1}(-1/\sqrt{7})+\pi}{4} + \frac{1}{4}n\pi
$$

The next example is a slight generalization of the procedure mentioned above.
Solve
$$
\cot x \cos^2 x = \frac{1}{2}\cot x
$$


Let \(u = \cot x\) and \(v = \cos x\), then
$$
u v^2 – \frac{1}{2}u = 0
$$
Now factor.
$$
u \left(v^2- \frac{1}{2} \right) = 0
$$
That is, either
$$
u = 0 \text{ or } v^2 -\frac{1}{2} = 0.
$$
That is, either
$$
\cot x = 0 \text{ or } \cos x = \pm \frac{1}{\sqrt{2}}.
$$
That is, either
$$
x = \pm\frac{\pi}{2} + 2 n \pi
$$
or
$$
x = \pm \frac{\pi}{4} + 2n \pi
$$
or
$$
x = \pm \frac{3\pi}{4} + 2n \pi
$$
Solve
$$
2\sin^2 x – \sin x – 1 = 0
$$


Let \(u = \sin x\). Then
$$
2u^2 – u – 1 = 0.
$$
So,
$$
(2u+1)(u-1) = 0
$$
Therefore, either
$$
u = \frac{-1}{2} \text{ or } u = 1.
$$

So, either
$$
\sin x = \frac{-1}{2} \text{ or } \sin x = 1.
$$

That is, either
$$
x = \frac{-\pi}{6} + 2n\pi
$$
or
$$
x = \frac{-5\pi}{6} + 2n \pi
$$
or
$$
x = \frac{\pi}{2} + 2n\pi
$$

Rewrite using a Single Trigonometric Function: If the trigonometric equation can be rewritten using a single trigonometric function, then rewrite it as so and apply the other strategies to solve.
Solve
$$
2\cos^2 x + 3\sin x -3 = 0
$$


Recall that \(\sin^2x + \cos^2 x = 1\), or \(\cos^2 x = 1-\sin^2x\).
So the given equation can be expressed as
$$
2(1-\sin^2 x) + 3 \sin x -3 =0
$$
or
$$
2\sin^2x -3 \sin x +1=0.
$$
Let \(u = \sin x.\) Then
$$
2u^2 – 3u +1 = 0
$$
or
$$
(2u-1)(u-1) = 0.
$$
That is,
$$
u = \frac{1}{2} \text{ or } u = 1.
$$
That is,
$$
\sin x = \frac{1}{2} \text{ or } \sin x = 1.
$$
That is, either
$$
x = \frac{\pi}{6} + 2n\pi
$$
or
$$
x = \frac{5\pi}{6} + 2n \pi
$$
or
$$
x = \frac{3\pi}{2} + 2n\pi
$$
Square Both Sides and Rewrite Using a Single Trigonometric Function: If squaring both sides of the equation allows you to rewrite the equation using a single trigonometric function, then do as such and apply the previous method to solve. In this method, squaring can expand the original solution set. You must check all solutions to ensure they are valid.
Find all solutions
$$
\cos x + 1 = \sin x
$$
on \([0, 2\pi) \).


E 5.3 Exercises