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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. 20 (page 465)

In Problems 11鈥�34, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$
$\tan (2 胃) = - 1$

Short Answer

Expert verified

The solution set is $\{\frac{3 蟺}{8}, \frac{7 蟺}{8}, \frac{11 蟺}{8}, \frac{15 蟺}{8}\}$ .

Step by step solution

01

Step 1. Given Information

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

$\tan (2 胃) = - 1$

02

Step 2. In the interval [0,2π), the tangent function equals -1 at 3π4

$2 胃$ must equal to $\frac{3 蟺}{4}$ .

$2 胃 = \frac{3 蟺}{4} + 蟺 n$

Divide by 2 on both side

$\frac{2}{2} 胃 = \frac{1}{2} (\frac{3 蟺}{4} + 蟺 n) 胃 = \frac{1}{2} 路 \frac{3 蟺}{4} + \frac{1}{2} 路蟺 n 胃 = \frac{3 蟺}{8} + \frac{蟺 n}{2}$

03

Step 3. The general formula is θ=3π8+πn2

So the value of given function in interval $[0, 2 蟺)$ is

$胃 = \frac{3 蟺}{8} + \frac{蟺脳 0}{2} 胃 = \frac{3 蟺}{8} + \frac{蟺脳 1}{2} 胃 = \frac{3 蟺}{8} + \frac{蟺脳 2}{2} 胃 = \frac{3 蟺}{8} + \frac{蟺脳 3}{2} 胃 = \frac{3 蟺}{8} 胃 = \frac{3 蟺}{8} + \frac{蟺}{2} 胃 = \frac{3 蟺}{8} + 蟺胃 = \frac{3 蟺}{8} + \frac{3 蟺}{2} 胃 = \frac{3 蟺}{8} 胃 = \frac{3 蟺 + 4 蟺}{8} 胃 = \frac{3 蟺 + 8 蟺}{8} 胃 = \frac{3 蟺 + 12 蟺}{8} 胃 = \frac{3 蟺}{8} 胃 = \frac{7 蟺}{8} 胃 = \frac{11 蟺}{8} 胃 = \frac{15 蟺}{8}$

So the solution set is $\{\frac{3 蟺}{8}, \frac{7 蟺}{8}, \frac{11 蟺}{8}, \frac{15 蟺}{8}\}$ .

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Most popular questions from this chapter

True or False: $\sin (- 胃) + \cos (- 胃) = \cos (胃) - \sin (胃)$

True or False: $\sin (- 胃) + \sin (胃) = 0$ for any value of theta

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

$\cos (\sin^{- 1} \frac{\sqrt{2}}{2})$

Establish each identity.

$\frac{1 - 2 \cos^{2} 胃}{\sin 胃 \cos 胃} = \tan 胃 - c o t 胃$

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

$\cos [\sin^{- 1} (- \frac{\sqrt{3}}{2})]$

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