Solving Trigonometric Equations

By the end of this lesson, you will be able to solve trig equations over specific intervals and in general form, using factoring, identities, and inverse trig functions.

Strategy Overview

1. Isolate the trig function (one trig function on one side)
2. If multiple trig functions: use identities to write in one function
3. Find all solutions in $[0, 2\pi)$: use reference angles and quadrant analysis
4. If general solution needed: add $2n\pi$ (sin/cos) or $n\pi$ (tan)

Example 1 — Isolate and Solve

Solve $2\cos x - \sqrt{3} = 0$ for $x \in [0, 2\pi)$.

$\cos x = \sqrt{3}/2$

Reference angle: $\pi/6$. Cosine positive in QI and QIV.

$x = \pi/6$ and $x = 11\pi/6$

Example 2 — Quadratic Type

Solve $2\sin^2 x - \sin x - 1 = 0$ for $x \in [0, 2\pi)$.

Let $u = \sin x$: $(2u+1)(u-1) = 0$

$\sin x = -1/2$: reference angle $\pi/6$; negative sine in QIII and QIV: $x = 7\pi/6, 11\pi/6$

$\sin x = 1$: $x = \pi/2$

Solutions: ${7\pi/6, 11\pi/6, \pi/2}$

Example 3 — Using an Identity

Solve $\sin x = \cos x$ for $x \in [0, 2\pi)$.

Divide by $\cos x$ (check $\cos x \neq 0$):

$\tan x = 1$; reference angle $\pi/4$; tangent positive in QI and QIII.

$x = \pi/4$ and $x = 5\pi/4$

Example 4 — Double Angle Equation

Solve $\sin(2x) = \cos x$ for $x \in [0, 2\pi)$.

$2\sin x\cos x = \cos x$

$\cos x(2\sin x - 1) = 0$

$\cos x = 0$: $x = \pi/2, 3\pi/2$

$\sin x = 1/2$: $x = \pi/6, 5\pi/6$

All four solutions: ${\pi/6, \pi/2, 5\pi/6, 3\pi/2}$

Example 5 — General Solution

Find the general solution to $\tan^2 x - \tan x = 0$.

$\tan x(\tan x - 1) = 0$

$\tan x = 0$: $x = n\pi$

$\tan x = 1$: $x = \pi/4 + n\pi$

General solution: $x = n\pi$ or $x = \pi/4 + n\pi$ for all integers $n$.

Example 6 — Equation with Period Change

Solve $\cos(2x) = 1/2$ for $x \in [0, 2\pi)$.

$2x = \pi/3 + 2n\pi$ or $2x = 5\pi/3 + 2n\pi$

$x = \pi/6 + n\pi$ or $x = 5\pi/6 + n\pi$