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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. 74 (page 504)

Solve the equation on the interval $0 鈮� 胃鈮� 2 蟺$ .
$\sin (2 胃) = \sqrt{2} \cos 胃$

Short Answer

Expert verified

On solving the equation, we get,

$胃鈭� \{\frac{蟺}{4}, \frac{蟺}{2}, \frac{3 蟺}{4}, \frac{3 蟺}{2}\}$

Step by step solution

01

Step 1. Given information.

Consider the given question,

$\sin (2 胃) = \sqrt{2} \cos 胃 0 鈮� 胃鈮� 2 蟺$

We know $\sin (2 胃) = 2 \sin 胃路 \cos 胃$ ,

$2 \sin 胃路 \cos 胃 = \sqrt{2} \cos 胃 2 \sin 胃路 \cos 胃 - \sqrt{2} \cos 胃 = 0 c os 胃 (2 \sin 胃 - \sqrt{2}) = 0$

Then,

localid="1646750340707" $\cos 胃 = 0$

Also,

$2 \sin 胃 - \sqrt{2} = 0 \sin 胃 = \frac{\sqrt{2}}{2} \sin 胃 = \frac{1}{\sqrt{2}}$

02

Step 2. Solve the equation for the given interval.

Consider the given question,

$\cos 胃 = 0 \sin 胃 = \frac{1}{\sqrt{2}} 0 鈮� 胃鈮� 2 蟺$

Solving $\cos 胃 = 0$ for the given interval,

$\cos 胃 = 0 胃 = \cos^{- 1} (0) 胃 = \frac{蟺}{2}, \frac{3 蟺}{2}$

Solving role="math" localid="1646750410065" $\sin 胃 = \frac{1}{\sqrt{2}}$ for the given interval,

$\sin 胃 = \frac{1}{\sqrt{2}} 胃 = \sin^{- 1} (\frac{1}{\sqrt{2}}) 胃 = \frac{蟺}{4}, \frac{3 蟺}{4}$

Therefore, the solution set is $胃鈭� \{\frac{蟺}{4}, \frac{蟺}{2}, \frac{3 蟺}{4}, \frac{3 蟺}{2}\}$ .

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Most popular questions from this chapter

Find the value of the number C:

$\frac{1}{2} \sin^{2} x + C = - \frac{1}{4} \cos (2 x)$

Multiply: $\frac{\sin (胃)}{1 + \cos (胃)} \frac{1 - \cos (胃)}{1 - \cos (胃)}$

Express the product $\cos 3 胃 \cos 5 胃$ as a sum containing only sines or only cosines.

Establish each identity.

$\frac{1 + \cos 胃 + \sin 胃}{1 + \cos 胃 - \sin 胃} = s e c 胃 + \tan 胃$

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

$\cot [\sin^{- 1} (- \frac{1}{2})]$

See all solutions

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