# Chapter 7 - Trigonometric Identities and Equations - 7.6 Trigonometric Equations - 7.6 Exercises - Page 720: 3

$\displaystyle \{\frac{7\pi}{6}, \ \ \frac{11\pi}{6}\}$

#### Work Step by Step

For a point on the unit circle, the coordinates are ($\cos x,\sin x)$, $x\in[0,2\pi)$ To solve the problem, find the points for which the $y$-coordinate is $-\displaystyle \frac{1}{2}$, ($\sin x$=$-\displaystyle \frac{1}{2}$) Using the given unit circle, we find such points in quadrant III, for $x=\displaystyle \frac{7\pi}{6}$ in quadrant IV, for $x=\displaystyle \frac{11\pi}{6}$ Solution set: $\displaystyle \{\frac{7\pi}{6}, \ \ \frac{11\pi}{6}\}$

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