Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 34


The solution set to this problem is $$\{90^\circ,270^\circ\}$$

Work Step by Step

$$\sin^2\theta\cos\theta=\cos\theta$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$\sin^2\theta\cos\theta=\cos\theta$$ Again, 2 different trigonometric function types would not help, so we need to find a way to eliminate either sine or cosine function, or separate them into different equations. $$\cos\theta-1\sin^2\theta\cos\theta=0$$ $$\cos\theta(1-\sin^2\theta)=0$$ Recall the identity: $1-\sin^2\theta=\cos^2\theta$ $$\cos^3\theta=0$$ $$\cos\theta=0$$ 2) Apply the inverse function: Over the interval $[0^\circ,360^\circ)$, examining the unit circle, it is found that there are 2 values of $\theta$ where $\cos\theta=0$, which are $\{90^\circ,270^\circ\}$ Therefore, $$\theta\in\{90^\circ,270^\circ\}$$ In other words, the solution set to this problem is $$\{90^\circ,270^\circ\}$$
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