Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Review Exercises - Page 291: 36


The solutions are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.

Work Step by Step

Take the square root of both sides to obtain: $\sin^2{x} = 1 \\\sqrt{\sin^2{x}}=\pm\sqrt1 \\\sin{x} = \pm 1$ Thus, $\sin{x} = -1$ or $\sin{x} = 1$. Solve each equation to obtain: \begin{array}{ccc} &\sin{x}=-1 &\text{or} &\sin{x} = 1 \\&x=\sin^{-1}{(-1)} &\text{or} &x=\sin^{-1}{(1)} \\&x=-\frac{\pi}{2} &\text{or} &x=\frac{\pi}{2} \end{array} Only $\frac{\pi}{2}$ is in the interval $[0, 2\pi)$. However, since the sine function is sinusoidal, note that $\sin{(\frac{3\pi}{2})}=-1$, which makes $\sin^2{(\frac{3\pi}{2})}=1$. Therefore, the solutions are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
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