## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 703: 32

#### Answer

The solution in the interval $[0,2\pi )$ is $\frac{\pi }{3}$.

#### Work Step by Step

We know that the period of the tangent function is $\pi$. In the interval $(0,\,\,\pi ]$, the only value for which the tangent function equals $\frac{\sqrt{3}}{3}$ is $\frac{\pi }{6}$. All the solutions to $\tan \frac{x}{2}=\frac{\sqrt{3}}{3}$ are given by: \begin{align} & \frac{x}{2}=\frac{\pi }{6}+n\pi \\ & x=\frac{\pi }{3}+2n\pi \end{align} Where, n is any integer. And the solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$. So, the equation is calculated by taking first $n$ as 0. It can be further simplified as follows. \begin{align} & x=\frac{\pi }{3}+2n\pi \\ & =\frac{\pi }{3}+\frac{0\times \pi }{3} \\ & =\frac{\pi }{3}+0 \\ & =\frac{\pi }{3} \end{align}

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