3.6 Trigonometric Equations

1. Core Methodology and General Solutions

To solve trigonometric equations, isolate the reference angle and apply periodic formulas to find all solutions across the domain.

For the forms $\sin x = \sin\theta$, $\cos x = \cos\theta$, and $\tan x = \tan\theta$, the general solutions are:

Function	Solution (Degrees)	Solution (Radians)
$\sin x = \sin\theta$	$x = \theta + 360^\circ k$ $x = (180^\circ - \theta) + 360^\circ k$	$x = \theta + 2k\pi$ $x = (\pi - \theta) + 2k\pi$
$\cos x = \cos\theta$	$x = \theta + 360^\circ k$ $x = -\theta + 360^\circ k$	$x = \theta + 2k\pi$ $x = -\theta + 2k\pi$
$\tan x = \tan\theta$	$x = \theta + 180^\circ k$	$x = \theta + k\pi$

Note: For $\sin x = 0$ or $\cos x = 0$, the solutions simplify to: $\sin x = 0 \implies x = k\pi$ and $\cos x = 0 \implies x = \dfrac{\pi}{2} + k\pi$.

Visualizing Trigonometric Solutions

Graphically, solving $\sin x = k$ means finding where the wave $y = \sin x$ intersects the line $y = k$.

2. Basic Domain Restrictions

To solve within a specific domain, find the general solution first, then substitute integer values for $k$ to find the solutions within the boundaries.

EXAMPLE 2

Solve $\sin x = \dfrac{1}{2}$ for $0^\circ \le x \le 360^\circ$:

The reference angle is $30^\circ$.

General solutions:
$x = 30^\circ + 360^\circ k$
$x = 150^\circ + 360^\circ k$

For $k = 0$, the solutions are $\mathbf{x = 30^\circ}$ and $\mathbf{x = 150^\circ}$.

EXAMPLE 3

Solve $\tan x = 1$ for $-180^\circ \le x \le 180^\circ$:

The reference angle is $45^\circ$.

General solution: $x = 45^\circ + 180^\circ k$.

Substitute integers for $k$:
For $k = 0$: $\mathbf{x = 45^\circ}$
For $k = -1$: $\mathbf{x = -135^\circ}$

3. Multiple Angle Adjustments

For multiple angles ($2x, 3x$, etc.), write the general solution before dividing to isolate $x$.

EXAMPLE 4

Solve $\sin 2x = \dfrac{\sqrt{3}}{2}$ for $0 \le x \le 2\pi$.

Reference equation: $\sin 2x = \sin\left(\dfrac{\pi}{3}\right)$.

Write the general solutions for $2x$ and solve for $x$:
$2x = \dfrac{\pi}{3} + 2k\pi \implies x = \dfrac{\pi}{6} + k\pi$
$2x = \dfrac{2\pi}{3} + 2k\pi \implies x = \dfrac{\pi}{3} + k\pi$

For $k = 0$ and $k = 1$, the solutions within the domain are:
$\mathbf{x = \dfrac{\pi}{6}, \dfrac{\pi}{3}, \dfrac{7\pi}{6}, \dfrac{4\pi}{3}}$.

EXAMPLE 5

Solve $\cos 3x = 0$ for $-180^\circ \le x \le 180^\circ$:

General solution: $3x = 90^\circ + 180^\circ k \implies x = 30^\circ + 60^\circ k$.

Substitute integers for $k$ to find the solutions:
$\mathbf{x = 30^\circ, 90^\circ, 150^\circ, -30^\circ, -90^\circ, -150^\circ}$.

EXAMPLE 6

Solve $\cos 2x = \dfrac{\sqrt{2}}{2}$ for $0 \le x \le 2\pi$:

General solutions are $2x = \dfrac{\pi}{4} + 2k\pi$ and $2x = -\dfrac{\pi}{4} + 2k\pi$.

Divide by 2 to isolate $x$: $x = \dfrac{\pi}{8} + k\pi$ and $x = -\dfrac{\pi}{8} + k\pi$.

The solutions within the domain are:
$\mathbf{x = \dfrac{\pi}{8}, \dfrac{9\pi}{8}, \dfrac{7\pi}{8}, \dfrac{15\pi}{8}}$.

EXAMPLE 7

Solve $\tan(x - \dfrac{\pi}{4}) = 1$ for $0 \le x \le 2\pi$:

Find the general solution using the reference angle $\dfrac{\pi}{4}$.

General solution: $x - \dfrac{\pi}{4} = \dfrac{\pi}{4} + k\pi \implies x = \dfrac{\pi}{2} + k\pi$.

The solutions within the domain are:
$\mathbf{x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}}$.

4. Reducible and Quadratic Trigonometric Forms

Equations with different terms or powers can be solved using factoring, substitution, or trigonometric identities.

EXAMPLE 8 (Factorization via Identities)

Solve $\sin 2x = \sin x$ for $0^\circ \le x \le 360^\circ$.

Use the double-angle identity: $2\sin x\cos x = \sin x$.

Move all terms to one side: $2\sin x\cos x - \sin x = 0$.

Factor the equation: $\sin x(2\cos x - 1) = 0$.

Set each factor to zero:
Path 1: $\sin x = 0 \implies \mathbf{x = 0^\circ, 180^\circ, 360^\circ}$.
Path 2: $\cos x = \dfrac{1}{2} \implies \mathbf{x = 60^\circ, 300^\circ}$.

EXAMPLE 9 (Direct Quadratic Form)

Solve $2\cos^2 x - 3\cos x + 1 = 0$ for $0 \le x \le \pi$.

Substitute $y = \cos x$ to get the quadratic equation $2y^2 - 3y + 1 = 0$.

Factor to find the roots: $(2y - 1)(y - 1) = 0 \implies y = 1 \text{ or } y = \dfrac{1}{2}$.

Substitute $\cos x$ back for $y$:
$\cos x = 1 \implies \mathbf{x = 0}$
$\cos x = \dfrac{1}{2} \implies \mathbf{x = \dfrac{\pi}{3}}$

EXAMPLE 10 (Mixed Term Quadratic)

Solve $3(1 - \cos x) = 2\sin^2 x$ for $0 \le x \le \pi$.

Use the Pythagorean identity to rewrite the equation in terms of cosine.

$$\begin{aligned} 3(1 - \cos x) &= 2(1 - \cos^2 x) \\ 3 - 3\cos x &= 2 - 2\cos^2 x \\ 2\cos^2 x - 3\cos x + 1 &= 0 \end{aligned}$$

This mirrors the structure from Example 9, yielding $\mathbf{x = 0}$ and $\mathbf{x = \dfrac{\pi}{3}}$.

EXAMPLE 11 (Combining Ratios)

Solve $\sqrt{3}\sin x = \cos x$ for $0 \le x \le 2\pi$.

Divide by $\cos x$ to rewrite the equation in terms of tangent.

This gives $\tan x = \dfrac{1}{\sqrt{3}}$.

The solutions within the domain are:
$\mathbf{x = \dfrac{\pi}{6}}$ and $\mathbf{x = \dfrac{7\pi}{6}}$.