Solving Trig Equations: General Solutions

By the end of this lesson, you will be able to find all solutions (general solutions) of trig equations over the reals, express them with the period parameter $n$, and solve equations requiring identities to simplify before solving.

The Big Idea

Trig equations have infinitely many solutions because trig functions are periodic. A general solution captures all solutions by expressing one complete period's solutions and adding the period parameter $n$ (an integer). This is the full solution — restricting to $[0, 2\pi)$ gives only a subset.

General Solution Formulas

$\sin\theta = k$:

$$\theta = \sin^{-1}(k) + 2n\pi \quad \text{and} \quad \theta = \pi - \sin^{-1}(k) + 2n\pi$$

$\cos\theta = k$:

$$\theta = \pm\cos^{-1}(k) + 2n\pi$$

$\tan\theta = k$:

$$\theta = \tan^{-1}(k) + n\pi$$

(Tangent has period $\pi$, so only one formula needed.)

Example 1 — Cosine Equation

Solve $\cos\theta = -\frac{1}{2}$ for all real $\theta$.

$\cos^{-1}(1/2) = \pi/3$

$\cos\theta = -1/2$: angles in QII and QIII.

$\theta = \pi - \pi/3 = 2\pi/3$ and $\theta = \pi + \pi/3 = 4\pi/3$ in $[0, 2\pi)$.

General solution: $\theta = \dfrac{2\pi}{3} + 2n\pi$ or $\theta = \dfrac{4\pi}{3} + 2n\pi$

Equivalently: $\theta = \pm\dfrac{2\pi}{3} + 2n\pi$

Example 2 — Tangent Equation

Solve $\tan\theta = \sqrt{3}$ for all real $\theta$.

$\tan^{-1}(\sqrt{3}) = \pi/3$

General solution: $\theta = \dfrac{\pi}{3} + n\pi$ for all integers $n$.

Example 3 — Linear in a Trig Function

Solve $2\sin(3x) - 1 = 0$.

$\sin(3x) = 1/2$

Let $u = 3x$: $\sin u = 1/2$ → $u = \pi/6 + 2n\pi$ or $u = 5\pi/6 + 2n\pi$

$3x = \pi/6 + 2n\pi \implies x = \pi/18 + 2n\pi/3$

$3x = 5\pi/6 + 2n\pi \implies x = 5\pi/18 + 2n\pi/3$

Example 4 — Quadratic-Type Trig Equation

Solve $2\cos^2\theta - \cos\theta - 1 = 0$.

Let $u = \cos\theta$: $2u^2 - u - 1 = (2u+1)(u-1) = 0$

$\cos\theta = -1/2$: general solution $\theta = \pm 2\pi/3 + 2n\pi$