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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 78 (page 496)

In Problems $69 - 78$ , solve each equation on the interval $0 鈮� 胃 < 2 蟺$
$\tan (2 胃) + 2 \cos 胃 = 0$

Short Answer

Expert verified

The solution set is $胃鈭� \{\frac{蟺}{2}, \frac{7 蟺}{6}, \frac{3 蟺}{2}, \frac{11 蟺}{6}\}$

Step by step solution

01

Given information

Given function $\tan (2 胃) + 2 \cos 胃 = 0$

02

Simplifying the equation

Simplifying, we get

$\tan (2 胃) + 2 \cos 胃 = 0 \frac{\sin (2 胃)}{\cos (2 胃)} + 2 \cos 胃 = 0 \sin (2 胃) + 2 \cos 胃 \cos (2 胃) = 0 2 \sin 胃 \cos 胃 + 2 \cos 胃 (1 - 2 \sin^{2} 胃) = 0$

03

Solving further

Solving, we get

$2 \cos 胃 (\sin 胃 + 1 - 2 \sin^{2} 胃) = 0 \cos 胃 (2 \sin^{2} 胃 - \sin 胃 - 1) = 0 \cos 胃 (2 \sin 胃 + 1) (\sin 胃 - 1) = 0 \cos 胃 = 0 \sin 胃 = - \frac{1}{2} 胃 = \frac{蟺}{2} 胃 = \frac{3 蟺}{2} 胃 = \frac{7 蟺}{6} 胃 = \frac{11 蟺}{6} 胃 = \frac{蟺}{2}$

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Most popular questions from this chapter

express each product as a sum containing only sines or only cosines

$\cos 3 蠎 \cos 4 蠎$

Show that $c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$

Establish each identity.

$\frac{1}{1 - \sin 胃} + \frac{1}{1 + \sin 胃} = 2 s e c^{2} 胃$

Show that $\cos \frac{蟺}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ and use it to find $\sin \frac{蟺}{16}, \cos \frac{蟺}{16}$ .

$\tan \frac{胃}{2} = \frac{1 - \cos 胃}{_____}$ .

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