Solving Equations Using Identities

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Transform non-factorable trigonometric equations using Pythagorean and double-angle formulas.

Solving Equations Using Identities

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

Substitute $\cos(2x) = 1 - 2\sin^2(x)$: $$ (1 - 2\sin^2(x)) + \sin^2(x) = 0 $$ $$ 1 - \sin^2(x) = 0 \implies \sin^2(x) = 1 \implies \sin(x) = \pm 1 $$ Sine is $\pm 1$ at: $$ x = \frac{\pi}{2}, \frac{3\pi}{2} $$

When is it necessary to use trigonometric identities before solving an equation?

Identities are required when an equation contains a mixture of different trigonometric functions (e.g., both sine and cosine) or different angle arguments (e.g., both $2x$ and $x$) that cannot be factored directly.

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

Substitute $\cos^2(x) = 1 - \sin^2(x)$: $$ 2(1 - \sin^2(x)) - \sin(x) - 1 = 0 $$ $$ -2\sin^2(x) - \sin(x) + 1 = 0 \implies (2\sin(x) - 1)(\sin(x) + 1) = 0 $$ 1. $\sin(x) = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}$ 2. $\sin(x) = -1 \implies x = \frac{3\pi}{2}$ Solutions: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$.

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

Use the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$: $$ 2\sin(x)\cos(x) - \cos(x) = 0 $$ $$ \cos(x)(2\sin(x) - 1) = 0 $$ 1. $\cos(x) = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}$ 2. $\sin(x) = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}$ Solutions: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$.

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

Use $\cos(2x) = 2\cos^2(x) - 1$: $$ 2\cos^2(x) - \cos(x) - 1 = 0 $$ $$ (2\cos(x) + 1)(\cos(x) - 1) = 0 $$ 1. $\cos(x) = -\frac{1}{2} \implies x = \frac{2\pi}{3}, \frac{4\pi}{3}$ 2. $\cos(x) = 1 \implies x = 0$ Solutions: $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$.

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

Substitute $\sec^2(x) = 1 + \tan^2(x)$: $$ 1 + \tan^2(x) - 2\tan(x) - 4 = 0 $$ $$ \tan^2(x) - 2\tan(x) - 3 = 0 \implies (\tan(x) - 3)(\tan(x) + 1) = 0 $$ 1. $\tan(x) = -1 \implies x = \frac{3\pi}{4}, \frac{7\pi}{4}$ 2. $\tan(x) = 3 \implies x = \arctan(3), \arctan(3) + \pi$.

Why must you check for extraneous solutions after squaring both sides of a trigonometric equation?

Squaring both sides can introduce false solutions because $A = B$ and $A = -B$ both result in $A^2 = B^2$. Every candidate solution must be plugged back into the original, unsquared equation.

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

Rearrange and square: $$ \sin(x) = 1 - \cos(x) \implies \sin^2(x) = 1 - 2\cos(x) + \cos^2(x) $$ Replace $\sin^2(x)$ with $1 - \cos^2(x)$: $$ 2\cos^2(x) - 2\cos(x) = 0 \implies 2\cos(x)(\cos(x) - 1) = 0 $$ Candidates: $x = 0, \frac{\pi}{2}, \frac{3\pi}{2}$. Check in original equation: - $x = 0: 0 + 1 = 1$ (Valid) - $x = \frac{\pi}{2}: 1 + 0 = 1$ (Valid) - $x = \frac{3\pi}{2}: -1 + 0 \ne 1$ (Extraneous) Solutions: $x = 0, \frac{\pi}{2}$.

Solving Equations Using Identities

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