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: 27


The solution set is $$\{0,\frac{\pi}{3},2\pi\}$$

Work Step by Step

$$\sin x=\sin2x$$ over interval $[0,2\pi)$ 1) In this case, only the interval for $x$, which is $[0,2\pi)$, is necessary, as you will see in step 2 that $\sin2x$ would be changed to a function of $x$ only. $$x\in[0,2\pi)$$ 2) Now consider back the equation $$\sin x=\sin2x$$ Here we see that $\sin x$ is a trigonometric function of $x$, but $\sin2x$ is that of $2x$. Thus it is essential to change $\sin2x$ to a trigonometric function of $x$ by using the identity $\sin2x=2\sin x\cos x$ $$\sin x=2\sin x\cos x$$ $$2\sin x\cos x-\sin x=0$$ $$\sin x(2\cos x-1)=0$$ $$\sin x=0\hspace{1cm}\text{or}\hspace{1cm}\cos x=\frac{1}{2}$$ For $\sin x=0$, over the interval $[0,2\pi)$, there are 2 values of $x$ whose $\sin x$ equals $0$, which are $\{0,2\pi\}$ For $\cos x=\frac{1}{2}$, over the interval $[0,2\pi)$, there is only one value of $x$ whose $\cos x$ equals $\frac{1}{2}$, which is $\{\frac{\pi}{3}\}$ Combining the solutions of 2 cases where $\sin x=0$ or $\cos x=\frac{1}{2}$, we end up with the solution set: $$x=\{0,\frac{\pi}{3},2\pi\}$$
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