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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. 69 (page 465)

In Problems 57鈥� 80, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$
$\tan 胃 = 2 \sin 胃$

Short Answer

Expert verified

The solution set of the given equation is $\frac{蟺}{3}, \frac{5 蟺}{3}$ .

Step by step solution

01

Step 1. Given Information

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

$\tan 胃 = 2 \sin 胃$

02

Step 2. The equation in its present form contains a tangent and sine.

We know that $\tan 胃 = \frac{\sin 胃}{\cos 胃}$

So $\frac{\sin 胃}{\cos 胃} = 2 \sin 胃$

Divide by $\sin 胃$ on both side

$\frac{\sin 胃}{\cos 胃} 路 \frac{1}{\sin 胃} = 2 \frac{\sin 胃}{\sin 胃} \frac{1}{\cos 胃} = 2$

03

Step 3. We know that secθ=1cosθ

So $\sec 胃 = 2$

Solving each equation in the interval $[0, 2 蟺]$ , we obtain

$胃 = \frac{蟺}{3}, \frac{5 蟺}{3}$

The solution set is $\frac{蟺}{3}, \frac{5 蟺}{3}$ .

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

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

Rewrite in terms of sine and cosine functions: $c o t (胃) s e c (胃)$

Show that $\tan^{- 1} (\frac{1}{v}) = \frac{蟺}{2} - \tan^{- 1} v$ , if $v > 0$

solve each equation on the interval

$0 鈮� 胃鈮� 2 蟺$

$\sin 胃 - \cos 胃 = - \sqrt{2}$

See all solutions

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