Trigonometry / Unit 7: Trigonometric Equations
7.2 Equations with Multiple Angles
Notes
For an equation like $\sin 2x = a$ on $[0, 2\pi)$, the angle $2x$ ranges over $[0, 4\pi)$, so go around the circle twice.
- List all values of $2x$ in $[0, 4\pi)$, then divide by 2.
- For $\displaystyle \sin\frac{x}{2} = a$, the angle $\displaystyle \frac{x}{2}$ lies in $[0, \pi)$, so fewer solutions appear.
- Example: $\displaystyle \cos 2x = \frac{1}{2}$ gives $\displaystyle 2x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}$, so there are 4 solutions.
Practice
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 $[0, 2\pi)$?
$2x$ runs through two full turns, and cosine equals $\displaystyle \frac{1}{2}$ twice per turn.
Solve $\displaystyle \sin\frac{x}{2} = \frac{1}{2}$ on $[0, 2\pi)$.
$\displaystyle \frac{x}{2}$ is in $[0, \pi)$, so $\displaystyle \frac{x}{2} = \frac{\pi}{6}$ or $\displaystyle \frac{5\pi}{6}$. Then $\displaystyle x = \frac{\pi}{3}$ or $\displaystyle \frac{5\pi}{3}$.
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