Factoring and Quadratic Trig Equations

Core Concepts

Techniques for solving higher-degree and multi-term trigonometric equations by factoring.

Factoring and Quadratic Trig Equations

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Why is dividing both sides of an equation like $\sin(x)\cos(x) = \sin(x)$ by $\sin(x)$ a fatal error?

Dividing by $\sin(x)$ loses all solutions where $\sin(x) = 0$. Instead, collect all terms on one side and factor: $$ \sin(x)(\cos(x) - 1) = 0 $$

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

Factor out $\sin(x)$: $$ \sin(x)(2\sin(x) - 1) = 0 $$ Set each factor to zero: 1. $\sin(x) = 0 \implies x = 0, \pi$ 2. $\sin(x) = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}$ Solution set: $$ x = 0, \frac{\pi}{6}, \pi, \frac{5\pi}{6} $$

Solve for $x \in [0, 2\pi)$: $$ 2\cos^2(x) - 3\cos(x) + 1 = 0 $$

Let $u = \cos(x)$, giving $2u^2 - 3u + 1 = (2u - 1)(u - 1) = 0$. 1. $\cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3}$ 2. $\cos(x) = 1 \implies x = 0$ Solution set: $$ x = 0, \frac{\pi}{3}, \frac{5\pi}{3} $$

Solve for $x \in [0, 2\pi)$: $$ 2\sin^2(x) + 3\sin(x) - 2 = 0 $$

Factor the quadratic: $$ (2\sin(x) - 1)(\sin(x) + 2) = 0 $$ 1. $2\sin(x) - 1 = 0 \implies \sin(x) = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}$ 2. $\sin(x) + 2 = 0 \implies \sin(x) = -2$, which has no real solutions. Solutions: $$ x = \frac{\pi}{6}, \frac{5\pi}{6} $$

Solve for $x \in [0, 2\pi)$: $$ \tan^2(x) - 3 = 0 $$

Take square roots: $$ \tan(x) = \pm\sqrt{3} $$ Reference angle is $\frac{\pi}{3}$. All four quadrants are included: $$ x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} $$

Solve for $x \in [0, 2\pi)$: $$ \sin(x)\tan(x) = \sin(x) $$

Move terms to one side and factor: $$ \sin(x)(\tan(x) - 1) = 0 $$ 1. $\sin(x) = 0 \implies x = 0, \pi$ 2. $\tan(x) = 1 \implies x = \frac{\pi}{4}, \frac{5\pi}{4}$ Solution set: $x = 0, \frac{\pi}{4}, \pi, \frac{5\pi}{4}$.

Solve for $x \in [0, 2\pi)$: $$ 4\cos^2(x) - 1 = 0 $$

Isolate $\cos^2(x)$: $$ \cos^2(x) = \frac{1}{4} \implies \cos(x) = \pm\frac{1}{2} $$ Reference angle is $\frac{\pi}{3}$. Since both signs are included: $$ x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} $$

Solve for $x \in [0, 2\pi)$ by grouping: $$ 2\sin(x)\cos(x) - \sqrt{3}\sin(x) - 2\cos(x) + \sqrt{3} = 0 $$

Group terms: $$ \sin(x)(2\cos(x) - \sqrt{3}) - 1(2\cos(x) - \sqrt{3}) = 0 $$ $$ (\sin(x) - 1)(2\cos(x) - \sqrt{3}) = 0 $$ 1. $\sin(x) = 1 \implies x = \frac{\pi}{2}$ 2. $\cos(x) = \frac{\sqrt{3}}{2} \implies x = \frac{\pi}{6}, \frac{11\pi}{6}$ Solution set: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{11\pi}{6}$.

Factoring and Quadratic Trig Equations

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