Precalculus / Unit 5: Trigonometric Identities and Equations
5.6 Equations with Multiple Angles and Identities
Notes
- For $\sin 2x = k$ on $[0, 2\pi)$, solve for $2x$ on $[0, 4\pi)$, then divide by 2. This doubles the number of solutions.
- Use a double-angle identity to get a single trig function, then factor. Never divide by a trig function that could be zero; factor it out.
- Example: $\sin 2x = \sin x \Rightarrow \sin x(2\cos x - 1) = 0$, so $\displaystyle x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$.
Practice
Solve $\sin 2x = \sin x$ on $[0, 2\pi)$.
$\sin x(2\cos x - 1) = 0$ gives $\sin x = 0$ or $\displaystyle \cos x = \frac{1}{2}$.
Solve $\displaystyle \sin 2x = \frac{\sqrt{3}}{2}$ on $[0, 2\pi)$.
$\displaystyle 2x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}$; divide each by 2.
How many solutions does $\displaystyle \cos 2x = \frac{1}{2}$ have on $[0, 2\pi)$?
$\displaystyle 2x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}$, giving four values of $x$.
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