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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. 44 (page 501)

Solve each equation on the interval
$0 鈮� 胃鈮� 2 蟺$
$\cos (2 胃) + \cos (4 胃) = 0$

Short Answer

Expert verified

$\{\frac{蟺}{6}, \frac{蟺}{2}, \frac{5 蟺}{6}, \frac{7 蟺}{6}, \frac{3 蟺}{2}, \frac{11 蟺}{6}\}$

Step by step solution

01

Step 1. Given information

We have been given $\cos (2 胃) + \cos (4 胃) = 0$ .

We have to Solve the equation

02

Step 2. Simplify the expression

We know,

$\cos (4 胃) = \cos (2 路 2 胃)$

and $\cos (2 x) = 2 \cos^{2} (x) - 1$

Therefore,

role="math" localid="1646549911824" $\cos (2 胃) + \cos (4 胃) = 0 \cos (2 胃) + 2 \cos^{2} (2 胃) + 1 = 0 (2 \cos (2 胃) - 1) (\cos (2 胃) + 1) = 0$

03

Step 3. Solve for each expression

Solve for $(2 \cos (2 胃) - 1)$

Let $x = 2 胃$ ,

$2 \cos x - 1 = 0 \cos x = \frac{1}{2}$

For this x must be equal to $\frac{蟺}{3} + k 蟺 or \frac{5 蟺}{3} + 办蟺$

Solve for $(\cos (2 胃) + 1)$

Let $x = 2 胃$

$\cos x + 1 = 0 \cos x = - 1$

For this, x must be equal to $蟺 + 办蟺$

Hence solution for this expression will be

$\{\frac{蟺}{6}, \frac{蟺}{2}, \frac{5 蟺}{6}, \frac{7 蟺}{6}, \frac{3 蟺}{2}, \frac{11 蟺}{6}\}$

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Most popular questions from this chapter

Use the fact that

$\cos \frac{蟺}{12} = \frac{1}{4} (\sqrt{6} + \sqrt{2})$

to find $\sin \frac{蟺}{24}, \cos \frac{蟺}{24}$ .

In Problems 9鈥�36, find the exact value of each expression.

$\cos^{- 1} (\sin \frac{5 蟺}{4})$

Establish each identity.

$\frac{\sin^{3} 胃 + \cos^{3} 胃}{\sin 胃 + \cos 胃} = 1 - \sin 胃 \cos 胃$

Show that $\sin (\sin^{- 1} (v) + \cos^{- 1} (v)) = 1$

$\cos (- 胃) - \cos (胃)$

See all solutions

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