## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Quiz (Sections 6.1-6.3) - Page 282: 9

#### Answer

The solution set is $$\{\frac{\pi}{6}, \frac{2\pi}{3}\}$$

#### Work Step by Step

$$3\cot2x-\sqrt3=0$$ over the interval $[0,2\pi)$ We carry out algebraic operations here normally as any equations: $$\cot2x=\frac{\sqrt3}{3}$$ Now we can solve for $2x$. Over the interval $[0,2\pi)$, there are 2 values of $2x$ with which $\cot2x=\frac{\sqrt3}{3}$, which are $\frac{\pi}{3}$ and $\frac{4\pi}{3}$. Therefore, $$2x=\{\frac{\pi}{3},\frac{4\pi}{3}\}$$ That means, $$x=\{\frac{\pi}{6}, \frac{2\pi}{3}\}$$ This is the solution set of the problem.

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