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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. 16 (page 465)

In Problems 11鈥�34, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$
$4 \cos^{2} 胃 - 3 = 0$

Short Answer

Expert verified

The solution set is $\{\frac{蟺}{6}, \frac{5 蟺}{6}, \frac{7 蟺}{6} \frac{11 蟺}{6}\}$ .

Step by step solution

01

Step 1. Given Information

In the given problem we have to solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$

$4 \cos^{2} 胃 - 3 = 0$

02

Step 2. Firstly solving the equation

Add 3 on both side

$4 \cos^{2} 胃 - 3 + 3 = 0 + 3 4 \cos^{2} 胃 = 3$

Divide by 4 on both side

$\frac{4}{4} \cos^{2} 胃 = \frac{3}{4} \cos^{2} 胃 = \frac{3}{4}$

Taking square root on both side

$\cos 胃 = 卤 \sqrt{\frac{3}{4}} \cos 胃 = 卤 \frac{\sqrt{3}}{2}$

$\cos 胃 = \frac{\sqrt{3}}{2} and \cos 胃 = - \frac{\sqrt{3}}{2}$

03

Step 3. After solving the equation cosθ=32 and cosθ=-32

The period of the cosine function is $2 蟺$ . In the interval $[0, 2 蟺)$ , there are two angles $胃$ for which

$\cos 胃 = \frac{\sqrt{3}}{2} : 胃 = \frac{蟺}{6} and 胃 = \frac{11 蟺}{6}$

04

Step 4. The period of the cosine function cosθ=-32 is 2π.

The period of the cosine function is $2 蟺$ . In the interval $[0, 2 蟺)$ , there are two angles $胃$ for which role="math" localid="1646583425702" $\cos 胃 = - \frac{\sqrt{3}}{2} : 胃 = \frac{5 蟺}{6} and 胃 = \frac{7 蟺}{6}$

The solution set is $\{\frac{蟺}{6}, \frac{5 蟺}{6}, \frac{7 蟺}{6} \frac{11 蟺}{6}\}$ .

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Most popular questions from this chapter

Express the product $\sin (4 胃) \sin (2 胃)$ as a sum containing only sines or only cosines.

The distance $d$ from the point $(2, - 3)$ to the point $(5, 1)$ is ______.

Multiply and simplify: $\frac{(1 + \tan (胃)) (1 + \tan (胃)) - s e c (胃)}{\tan (胃)}$

Establish each identity.

$s e c 胃 - \cos 胃 = \sin 胃 \tan 胃$

Establish each identity.

$\frac{{(2 \cos^{2} 胃 - 1)}^{2}}{\cos^{4} 胃 - \sin^{4} 胃} = 1 - 2 \sin^{2} 胃$

See all solutions

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