Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.1 - Solving Trigonometric Equations - 6.1 Problem Set - Page 326: 34

Answer

(a) $\theta=30^{\circ}+360^{\circ}k,60^{\circ}+360^{\circ}k,120^{\circ}+360^{\circ}k,150^{\circ}+360^{\circ}k,$ where k is any integer. (b) $\theta=30^{\circ},60^{\circ},120^{\circ},150^{\circ}.$

Work Step by Step

(a) $(2sin\theta-\sqrt 3)\times(2\sin\theta-1)=0$ Therefore, either $\sin\theta=\frac{\sqrt 3}{2} or \frac{1}{2}$ For, $\sin\theta=\frac{\sqrt 3}{2}$, $\theta=\sin^{-1}(\frac{\sqrt 3}{2})$, Since, $\sin\theta$ is positive in both quadrants 1 and 2, $\theta=60^{\circ} \,\,or \,\,180^{\circ}-60^{\circ}=120^{\circ},$ To find all values of $\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of sine function is $360^{\circ}$. So, $\theta=60^{\circ}+360^{\circ}k\,\,or \,\,120^{\circ}+360^{\circ}k$ For, $\sin\theta=\frac{1}{2}$, $\theta=\sin^{-1}(\frac{1}{2})$, Since, $\sin\theta$ is positive in both quadrants 1 and 2, $\theta=30^{\circ} \,\,or \,\,180^{\circ}-30^{\circ}=150^{\circ},$ To find all values of $\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of sine function is $360^{\circ}$. So, $\theta=30^{\circ}+360^{\circ}k\,\,or \,\,150^{\circ}+360^{\circ}k$ So, overall general solution of $\theta=30^{\circ}+360^{\circ}k,\,60^{\circ}+360^{\circ}k,\,120^{\circ}+360^{\circ}k,\,150^{\circ}+360^{\circ}k.$ (b) Now, if $0^{\circ}\leq\theta\lt360^{\circ}$, we get the solutions by substituting k=0. So, the required values of $\theta$ are $30^{\circ},60^{\circ},120^{\circ},150^{\circ}.$
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