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: 38


$\left\{0.7297, 2.4116, \frac{\pi}{2}\right\}$

Work Step by Step

Let $u=\sin{x}$ Replacing $\sin{x}$ by $u$ gives: $3u^2-5u+2=0$ Factor the trinomial to obtain: $(3u-2)(u-1)=0$ use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} &3u-2=0 &\text{or} &u-1=0 \\&3u=2 &\text{or} &u=1 \\&u=\frac{2}{3} &\text{or} &u=1 \end{array} Substitute $u$ with $\sin{x}$ to obtain: \begin{array}{ccc} &\sin{x} = \frac{2}{3} &\text{or} &\sin{x}=1 \\&x=\sin^{-1}{(\frac{2}{3})} &\text{or} &x=\sin^{-1}{(1)} \\&x=0.7297276562 &\text{or} &x=\frac{\pi}{2} \end{array} Note that for an angle $x$ in Quadrant I like $0.7297276562$, the value of $\sin{x}=\sin{(\pi-x)}$. Thus, $\sin{(0.7297276562)}=\sin{(\pi-0.7297276562)}$. $\pi-0.7297276562=2.411592654\approx 2.4116$. Therefore, the solutions to the given equation, rounded-off to four decimal places when necessary, are: $\left\{0.7297, 2.4116, \frac{\pi}{2}\right\}$
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