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: 25


The equation has the solution set: $$\{\frac{\pi}{2},\frac{3\pi}{2}\}$$

Work Step by Step

$$\sin\frac{x}{2}=\sqrt2-\sin\frac{x}{2}$$ over interval $[0,2\pi)$ 1) Find corresponding interval for $\frac{x}{2}$ Interval $[0,2\pi)$ can be written as $$0\le x\lt2\pi$$ That means, for $\frac{x}{2}$, the interval would be $$0\le\frac{x}{2}\lt\pi$$ or $$\frac{x}{2}\in[0,\pi)$$ 2) Now consider back the equation $$\sin\frac{x}{2}=\sqrt2-\sin\frac{x}{2}$$ $$2\sin\frac{x}{2}=\sqrt2$$ $$\sin\frac{x}{2}=\frac{\sqrt2}{2}$$ Over the interval $[0,\pi)$, there are 2 values whose $\sin$ equals $\frac{\sqrt2}{2}$, which are $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, meaning that $$\frac{x}{2}=\{\frac{\pi}{4},\frac{3\pi}{4}\}$$ So $$x=\{\frac{\pi}{2},\frac{3\pi}{2}\}$$
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