## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 705: 141

#### Answer

See explanations.

#### Work Step by Step

The sine functions in the two equations have different periodicity. For $sin(x)=\frac{1}{2}$, the period is $2\pi$ and there are two solutions in $[0,2\pi$. While for $sin(x)=\frac{1}{2}$, because the period is $\pi$, the number of solutions will double to four.

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