Trigonometry 7th Edition

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

Chapter 6 - Section 6.2 - More on Trigonometric Equations - 6.2 Problem Set - Page 332: 5

Answer

$30^{\circ}, 330^{\circ}$

Work Step by Step

Given $\sqrt 3\sec\theta$=2 .....(1) As we know that $\sec\theta$ is the reciprocal of $\cos\theta$ so we can write as $\sec\theta = \frac{1}{\cos\theta}$ .....(2) Substitute (2) in equation (1) we get $\sqrt 3(\frac{1}{\cos\theta})$=2 $\sqrt 3=2\cos\theta$ On dividing both the sides by 2 $\frac{\sqrt 3}{2}= \cos\theta$ $\theta = 30^{\circ}$ cosine function is positive in quadrant 1 & 4 $\theta=30^{\circ} $ or $\theta^{\circ} = 360^{\circ}-30$ $\theta=30^{\circ} $ or $\theta^{\circ} = 330^{\circ}$
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