## Precalculus (6th Edition)

Published by Pearson

# Chapter 7 - Trigonometric Identities and Equations - 7.6 Trigonometric Equations - 7.6 Exercises: 34

#### Answer

$\{90^{o}, 270^{o}\}$

#### Work Step by Step

$\sin^{2}\theta\cos\theta=\cos\theta\qquad/-\cos\theta$ $\sin^{2}\theta\cos\theta-\cos\theta=0$ $\cos\theta(\sin^{2}\theta-1)=0$ ... zero factor principle ... 1. $\quad\cos\theta=0$ or 2.$\quad \sin^{2}\theta-1=0$ $1.\quad\cos\theta=0$ for $\theta=90^{o}, 270^{o}$ $2.\quad \sin^{2}\theta=1\quad/\pm\sqrt{(...)}$ $\sin\theta=\pm 1$ $\theta=90^{o}, 270^{o}$ Solution set = $\{90^{o}, 270^{o}\}$

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