3.6 Trigonometric Equations

1. Core Methodology and General Solutions

Determining the solutions to base trigonometric equations requires isolating a primary reference angle, then applying established cyclical formulas to map every valid intersection point across an unbounded domain.

Given the standard forms $\sin x = \sin\theta$, $\cos x = \cos\theta$, and $\tan x = \tan\theta$, the complete mathematical sets of valid solutions are generated by:

Function	General Solution (Degrees)	General 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 specialized extreme limits such as $\sin x = 0$ or $\cos x = 0$, the opposing bounds collapse into unified linear expressions: $\sin x = 0 \Rightarrow x = k\pi$, and $\cos x = 0 \Rightarrow x = \frac{\pi}{2} + k\pi$.

EXAMPLE 2 & 3 (Basic Domain Restrictions)

Trigonometric equations are universally evaluated by locating the infinite general bounds first, followed by filtering discrete integer multipliers ($k$) to lock solutions inside the requested boundaries.

Solve $\sin x = \frac{1}{2}$ for $0^\circ \le x \le 360^\circ$:
The foundational reference angle equates to $30^\circ$.
General models: $x = 30^\circ + 360^\circ k$ and $x = 150^\circ + 360^\circ k$.
Isolating the designated bounds yields exactly $\mathbf{x = 30^\circ}$ and $\mathbf{x = 150^\circ}$.

Solve $\tan x = 1$ for $-180^\circ \le x \le 180^\circ$:
The baseline reference angle equates to $45^\circ$.
General model: $x = 45^\circ + 180^\circ k$.
Injecting valid integers calculates $\mathbf{x = 45^\circ}$ ($k=0$) and $\mathbf{x = -135^\circ}$ ($k=-1$).

2. Multiple Angle Adjustments

Equations manipulating compressed or stretched angular domains ($2x, 3x$, etc.) must adhere strictly to establishing the entire general form algebraically before division isolation occurs.

EXAMPLE 4

Solve the relation $\sin 2x = \frac{\sqrt{3}}{2}$ restricting the domain to $0 \le x \le 2\pi$.

Determine principal state: $\sin 2x = \sin\frac{\pi}{3}$.

Draft general sequences based explicitly on the combined $2x$ term:
$2x = \frac{\pi}{3} + 2k\pi \Rightarrow x = \frac{\pi}{6} + k\pi$
$2x = \frac{2\pi}{3} + 2k\pi \Rightarrow x = \frac{\pi}{3} + k\pi$

Extracting iterations within the bound $2\pi$ generates four targets: $\mathbf{x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}}$.

EXAMPLE 5 & 6 (Zeroes and Nested Variables)

Solve $\cos 3x = 0$ for $-180^\circ \le x \le 180^\circ$:
General reduction: $3x = 90^\circ + 180^\circ k \Rightarrow x = 30^\circ + 60^\circ k$.
Testing integer mappings identifies exactly: $\mathbf{x = 30^\circ, 90^\circ, 150^\circ, -30^\circ, -90^\circ, -150^\circ}$.

Solve $\cos 2x = \frac{\sqrt{2}}{2}$ for $0 \le x \le 2\pi$:
General sets formulate to $2x = \frac{\pi}{4} + 2k\pi$ and $2x = -\frac{\pi}{4} + 2k\pi$.
Algebraic division scales the models to $x = \frac{\pi}{8} + k\pi$ and $x = -\frac{\pi}{8} + k\pi$.
Valid bounding identifies strictly: $\mathbf{x = \frac{\pi}{8}, \frac{9\pi}{8}, \frac{7\pi}{8}, \frac{15\pi}{8}}$.

3. Reducible and Quadratic Trigonometric Forms

When trigonometric relations involve combinations of differing structural components, algebraic factoring, substitution rules, or identity translations are necessary.

EXAMPLE 8 (Factorization via Identities)

Solve the equality $\sin 2x = \sin x$ mapped within $0^\circ \le x \le 360^\circ$.

Applying the double-angle identity expands the function: $2\sin x\cos x = \sin x$.

Transfer components to construct a zeroed structure: $2\sin x\cos x - \sin x = 0$.

Factorizing logically generates: $\sin x(2\cos x - 1) = 0$.

This establishes two separate evaluation pathways:
Path 1: $\sin x = 0 \Rightarrow \mathbf{x = 0^\circ, 180^\circ, 360^\circ}$.
Path 2: $\cos x = 1/2 \Rightarrow \mathbf{x = 60^\circ, 300^\circ}$.

EXAMPLE 9 & 10 (Quadratic Substitution)

Direct Quadratic Form (Example 9): Solve $2\cos^2 x - 3\cos x + 1 = 0$ for $0 \le x \le \pi$.
A direct variable substitution mapping $y = \cos x$ exposes the standard quadratic $2y^2 - 3y + 1 = 0$.
Resolving the roots provides $y = 1$ and $y = 1/2$.
Translating backward calculates $\cos x = 1 \Rightarrow \mathbf{x = 0}$, and $\cos x = 1/2 \Rightarrow \mathbf{x = \frac{\pi}{3}}$.

Mixed Term Quadratic (Example 10): Solve $3(1 - \cos x) = 2\sin^2 x$ for $0 \le x \le \pi$.
Mixed trigonometric variables mandate homogenization. Substituting the Pythagorean identity alters the right side: $2(1 - \cos^2 x)$.
Expanding and shifting terms constructs: $3 - 3\cos x = 2 - 2\cos^2 x \Rightarrow 2\cos^2 x - 3\cos x + 1 = 0$.
This mirrors the precise structure resolved previously, yielding $\mathbf{x = 0}$ and $\mathbf{x = \frac{\pi}{3}}$.

EXAMPLE 11 (Combining Ratios)

Solve the structural format $\sqrt{3}\sin x = \cos x$ bounded to $0 \le x \le 2\pi$.

Equations formatted as $A\sin x = B\cos x$ compress efficiently via division into $A\tan x = B$.

Manipulating terms produces $\tan x = \frac{1}{\sqrt{3}}$.

The general angular mapping $x = \frac{\pi}{6} + k\pi$ calculates exact solutions at $\mathbf{x = \frac{\pi}{6}}$ and $\mathbf{x = \frac{7\pi}{6}}$.