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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. 18 (page 465)

In Problems 11鈥�34, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$
$\tan \frac{胃}{2} = \sqrt{3}$

Short Answer

Expert verified

The solution set is $\{\frac{2 蟺}{3}\}$

Step by step solution

01

Step 1. Given Information

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

$\tan \frac{胃}{2} = \sqrt{3}$

02

Step 2. In the interval [0,2π), the tangent function equals 3 at π3.

So, we know that $\frac{胃}{2}$ must equal $\frac{蟺}{3}$ .

To find these solutions, write the general formula that gives all the solutions.

$\frac{胃}{2} = \frac{蟺}{3} + 蟺苍$

Multiply with 2 on both side

$\frac{胃}{2} 脳 2 = 2 脳 (\frac{蟺}{3} + 蟺苍) 胃 = \frac{2 蟺}{3} + 2 蟺苍$

03

Step 3. The general formula is θ=2π3+2πn.

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

$胃 = \frac{2 蟺}{3} + 2 蟺苍$

role="math" localid="1646586437011" $胃 = \frac{2 蟺}{3} + 2 蟺脳 0 胃 = \frac{2 蟺}{3}$

So the solution set is $\{\frac{2 蟺}{3}\}$ .

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Most popular questions from this chapter

Show that $\cos (\sin^{- 1} (v) + \cos^{- 1} (v)) = 0$

Establish each identity.

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

True or False: $\sin^{2} (胃) = 1 - \cos^{2} (胃)$

Establish each identity.

$\frac{\sin 胃 + \cos 胃}{\cos 胃} - \frac{\sin 胃 - \cos 胃}{\sin 胃} = s e c 胃 c s c 胃$

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

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