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Solving Trigonometric Equations using Trigonometric Identities

An equation that contains trigonometric functions is called trigonometric equation .

Example:

$\begin{array}{l} \sin^{2} (x) + \cos^{2} (x) = 1 \\ 2 \sin (x) - 1 = 0 \\ \tan^{2} (2 x) - 1 = 0 \end{array}$

Solving Trigonometric Equations using Trigonometric Identities

Trigonometric identities are equations involving the trigonometric functions that are true for every value of the variables involved. You can use trigonometric identities along with algebraic methods to solve the trigonometric equations.

Extraneous Solutions

An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was exclude from the domain of the original equation.

When you solve trigonometric equations, sometimes you can obtain an equation in one trigonometric function by squaring each side, but this technique may produce extraneous solutions.

Example :

Find all the solutions of the equation in the interval $[0, 2 π)$ .

$2 \sin^{2} (x) = 2 + \cos (x)$

The equation contains both sine and cosine functions.

We rewrite the equation so that it contains only cosine functions using the Pythagorean Identity $\sin^{2} (x) = 1 - \cos^{2} (x)$ .

$\begin{array}{l} 2 (1 - \cos^{2} (x)) = 2 + \cos (x) \\ 2 - 2 \cos^{2} (x) = 2 + \cos (x) \\ - 2 \cos^{2} (x) - \cos (x) = 0 \\ 2 \cos^{2} (x) + \cos (x) = 0 \end{array}$

Factoring $\cos (x)$ we obtain, $\cos (x) (2 \cos (x) + 1) = 0$ .

By using zero product property , we will get $\cos (x) = 0$ , and $2 \cos (x) + 1 = 0$ which yields $\cos (x) = - \frac{1}{2}$ .

In the interval $[0, 2 π)$ , we know that $\cos (x) = 0$ when $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ . On the other hand, we also know that $\cos (x) = - \frac{1}{2}$ when $x = \frac{2 π}{3}$ and $x = \frac{4 π}{3}$ .

Therefore, the solutions of the given equation in the interval $[0, 2 π)$ are

${\frac{π}{2}, \frac{3 π}{2}, \frac{2 π}{3}, \frac{4 π}{3}}$ .