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 6.3 Trigonometric Equations II - 6.3 Exercises - Page 279: 34


The solution set is $$\{\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\}$$

Work Step by Step

$$\sin x\cos x=\frac{1}{4}$$ over interval $[0,2\pi)$ This equation features both sine and cosine of the same functions of $x$, and they are in the position of a multiplication. In fact, it recalls the identity: $$2\sin x\cos x=\sin 2x$$ $$\sin x\cos x=\frac{\sin 2x}{2}$$ Therefore, the given equation would now become $$\frac{\sin2x}{2}=\frac{1}{4}$$ $$\sin2x=\frac{1}{2}$$ Using the identity $2\sin x\cos x=\sin2x$ helps solve this exercise, since if we keep both sine and cosine, there is no known way to solve it (except probably graphing). 1) Find corresponding interval for $2x$: The interval for $x$ is $[0,2\pi)$. In other words, $$0\le x\lt2\pi$$ Thus, with $2x$, the inequality would be $$0\le2x\lt4\pi$$ So the interval for $2x$ is $[0,4\pi)$ 2) Now consider back the equation $$\sin2x=\frac{1}{2}$$ Over interval $[0,4\pi)$, there are 4 values whose $\sin$ equals $\frac{1}{2}$, which are $\{\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\}$ Therefore, $$2x\in\{\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\}$$ That means $$x\in\{\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\}$$
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