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: 39


The solution set is $$\{\frac{2\pi}{3}+2n\pi,\frac{4\pi}{3}+2n\pi,n\in Z\}$$

Work Step by Step

$$2\sqrt3\sin\frac{x}{2}=3$$ 1) Solve the equation over the interval $[0,2\pi)$ The interval for $x$ is $[0,2\pi)$ As a result, the interval for $\frac{x}{2}$ is $[0,\pi)$ $$2\sqrt3\sin\frac{x}{2}=3$$ $$\sin\frac{x}{2}=\frac{3}{2\sqrt3}=\frac{\sqrt3}{2}$$ Over the interval $[0,\pi)$, there are 2 values of $\frac{x}{2}$ where $\sin\frac{x}{2}=\frac{\sqrt3}{2}$, which are $\{\frac{\pi}{3},\frac{2\pi}{3}\}$ Therefore, $$\frac{x}{2}=\{\frac{\pi}{3},\frac{2\pi}{3}\}$$ We would stop here and not solve for $x$. 2) Solve the equation for all solutions Sine function has period $2\pi$, so we would add $2\pi$ to all solutions found in part 1) for $\frac{x}{2}$. $$\frac{x}{2}=\{\frac{\pi}{3}+2n\pi,\frac{2\pi}{3}+2n\pi, n\in Z\}$$ Thus, $$x=\{\frac{2\pi}{3}+2n\pi,\frac{4\pi}{3}+2n\pi,n\in Z\}$$
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