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: $\sin x = 0 \implies x = k\pi$, and $\cos x = 0 \implies x = \dfrac{\pi}{2} + k\pi$.

Visualizing Trigonometric Solutions

Graphically, solving a trigonometric equation like $\sin x = k$ is equivalent to finding the intersection points between the wave function $y = \sin x$ and the horizontal line $y = k$.

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.

EXAMPLE 2

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

The foundational reference angle equates to $30^\circ$.

General models:
$x = 30^\circ + 360^\circ k$
$x = (180^\circ - 30^\circ) + 360^\circ k \implies x = 150^\circ + 360^\circ k$

Isolating the designated bounds ($k=0$) yields exactly $\mathbf{x = 30^\circ}$ and $\mathbf{x = 150^\circ}$.

EXAMPLE 3

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:
For $k = 0$: $\mathbf{x = 45^\circ}$
For $k = -1$: $x = 45^\circ - 180^\circ \implies \mathbf{x = -135^\circ}$

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

Determine principal state: $\sin 2x = \sin\left(\dfrac{\pi}{3}\right)$.

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

Extracting iterations within the bound $2\pi$ ($k=0$ and $k=1$) generates four targets:
$\mathbf{x = \dfrac{\pi}{6}, \dfrac{\pi}{3}, \dfrac{7\pi}{6}, \dfrac{4\pi}{3}}$.

EXAMPLE 5

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

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

Testing integer mappings ($k = 0, 1, 2, -1, -2, -3$) identifies exactly:
$\mathbf{x = 30^\circ, 90^\circ, 150^\circ, -30^\circ, -90^\circ, -150^\circ}$.

EXAMPLE 6

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

General sets formulate to $2x = \dfrac{\pi}{4} + 2k\pi$ and $2x = -\dfrac{\pi}{4} + 2k\pi$.

Algebraic division scales the models to $x = \dfrac{\pi}{8} + k\pi$ and $x = -\dfrac{\pi}{8} + k\pi$.

Valid bounding ($k=0, 1$ for the first; $k=1, 2$ for the second) identifies strictly:
$\mathbf{x = \dfrac{\pi}{8}, \dfrac{9\pi}{8}, \dfrac{7\pi}{8}, \dfrac{15\pi}{8}}$.

EXAMPLE 7

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

Isolate the internal angle structure based on the reference angle $\dfrac{\pi}{4}$.

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

Extracting valid solutions within the given boundary yields:
$\mathbf{x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}}$.

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: $\sin x = 0 \implies \mathbf{x = 0^\circ, 180^\circ, 360^\circ}$.
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 $0 \le x \le \pi$.

A direct variable substitution mapping $y = \cos x$ exposes the standard quadratic $2y^2 - 3y + 1 = 0$.

Factorizing $(2y - 1)(y - 1) = 0$ resolves the roots providing $y = 1$ and $y = \dfrac{1}{2}$.

Translating backward calculates:
$\cos x = 1 \implies \mathbf{x = 0}$
$\cos x = \dfrac{1}{2} \implies \mathbf{x = \dfrac{\pi}{3}}$

EXAMPLE 10 (Mixed Term Quadratic)

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

Mixed trigonometric variables mandate homogenization. Substituting the Pythagorean identity alters the right side.

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

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

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

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