## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 280: 47

#### Answer

The solution set is $$\{11.8^\circ+n180^\circ, 78.2^\circ+n180^\circ, n\in Z\}$$

#### Work Step by Step

$$2-\sin2\theta=4\sin2\theta$$ $$5\sin2\theta=2$$ $$\sin2\theta=\frac{2}{5}$$ 1) First, we solve the equation over the interval $[0^\circ,360^\circ)$ For $\sin2\theta=\frac{2}{5}$, we have $$2\theta=\sin^{-1}\frac{2}{5}\approx23.6^\circ$$ Also, over the interval $[0^\circ,360^\circ)$, there is one more value of $2\theta$ where $\sin2\theta=\frac{2}{5}$, which is $2\theta=180^\circ-23.6\approx156.4^\circ$ Therefore, $$2\theta=\{23.6^\circ,156.4^\circ\}$$ 2) Solve the equation for all solutions Sine function has period $360^\circ$, so we would add $360^\circ$ to all solutions found in part 1) for $2\theta$. $$2\theta=\{23.6^\circ+n360^\circ,156.4^\circ+n360^\circ, n\in Z\}$$ $$\theta=\{11.8^\circ+n180^\circ, 78.2^\circ+n180^\circ, n\in Z\}$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.