Answered: Find all radian solutions using exact…

Trigonometry (MindTap Course List)

8th Edition

ISBN:9781305652224

Author:Charles P. McKeague, Mark D. Turner

Publisher:Charles P. McKeague, Mark D. Turner

Chapter6: Equations

Section6.3: Trigonometric Equations Involving Multiple Angles

Problem 35PS

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Question

Find all radian solutions using exact values only.

2 cos x - sec x +tan x = 0

Expert Solution

Step 1

$G i v e n e q u a t i o n - 2 \cos x - s e c x + \tan x = 0 2 \cos x - \frac{1}{\cos x} + \frac{\sin x}{\cos x} = 0 \frac{2 \cos x (\cos x) - 1 + \sin x}{\cos x} = 0 \frac{2 \cos^{2} x - 1 + \sin x}{\cos x} = 0 2 \cos^{2} x - 1 + \sin x = 0 2 (1 - \sin^{2} x) - 1 + \sin x = 0 2 - 2 \sin^{2} x - 1 + \sin x = 0 - 2 \sin^{2} x + \sin x + 1 = 0 2 \sin^{2} x - \sin x - 1 = 0 . . . . . . . . . . . . . . . . (1)$

Step 2

$E q u a t i o n (1) i s a q u a d r a t i c e q u a t i o n o f t h e f o r m, a x^{2} + b x + c = 0, T h e n u \sin g f o r m u l a, x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} N o w, \sin x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (2) (- 1)}}{2 (2)} \sin x = \frac{1 \pm \sqrt{1 + 8}}{4} \sin x = \frac{1 \pm \sqrt{9}}{4} \sin x = \frac{1 \pm 3}{4} \sin x = \frac{1 + 3}{4}, \frac{1 - 3}{4} \sin x = \frac{4}{4}, \frac{- 2}{4} \sin x = 1, - \frac{1}{2} W e h a v e g o t t w o v a l u e s o f \sin x .$

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