Q. 73 In Problems 69-78, solve each eq... [FREE SOLUTION]

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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. 73 (page 496)

In Problems $69 - 78$ , solve each equation on the interval $0 鈮� 胃 < 2 蟺$ .
$\sin (2 胃) + \sin (4 胃) = 0$ .

Short Answer

Expert verified

The solution set for the equation $\sin (2 胃) + \sin (4 胃) = 0$ is, $\{0, \frac{蟺}{3}, \frac{蟺}{2}, 蟺, \frac{2 蟺}{3}, \frac{4 蟺}{3}, \frac{3 蟺}{2}, \frac{5 蟺}{3}\}$ .

Step by step solution

Step 1Given equation is,

$\sin (2 胃) + \sin (4 胃) = 0$ .

The double angle for $\sin (2 胃)$ is, $\sin (2 胃) = 2 \sin 胃 \cos 胃$ and for $\cos (2 胃)$ is, $\cos (2 胃) = 2 \cos^{2} 胃 - 1$ .

Substitute the known values in the given equation.

$\sin (2 胃) + 2 \sin (2 胃) \cos (2 胃) = 0 \sin (2 胃) (1 + 2 \cos (2 胃)) = 0 [2 \sin 胃 \cos 胃] (1 + 2 (2 \cos^{2} 胃 - 1)) = 0 [2 \sin 胃 \cos 胃] (1 + 4 \cos^{2} 胃 - 2) = 0 [2 \sin 胃 \cos 胃] (4 \cos^{2} 胃 - 1) = 0$

Step 2 Apply the zero product property.

$[2 \sin 胃 \cos 胃] = 0 o r (4 \cos^{2} 胃 - 1) = 0 \sin 胃 \cos 胃 = 0 o r 4 \cos^{2} 胃 = 1 \sin 胃 \cos 胃 = 0 o r \cos^{2} 胃 = \frac{1}{4} \sin 胃 \cos 胃 = 0 o r \cos 胃 = 卤 \frac{1}{2} \sin 胃 = 0 o r \cos 胃 = 0 o r \cos 胃 = 卤 \frac{1}{2}$

Step 3 Solving the three equations in the interval [0,2π) we get 7 values for θ.

$胃 = 0, \frac{蟺}{3}, \frac{蟺}{2}, \frac{2 蟺}{3}, 蟺, \frac{4 蟺}{3}, \frac{3 蟺}{2}, \frac{5 蟺}{3}$ .

Therefore the solution set is, $\{0, \frac{蟺}{3}, \frac{蟺}{2}, \frac{2 蟺}{3}, 蟺, \frac{4 蟺}{3}, \frac{3 蟺}{2}, \frac{5 蟺}{3}\}$ .

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91影视

In Problems $69 - 78$ , solve each equation on the interval $0 鈮� 胃 < 2 蟺$ .
$\sin (2 胃) + \sin (4 胃) = 0$ .

Short Answer

Step by step solution

Step 1Given equation is,

Step 2 Apply the zero product property.

Step 3 Solving the three equations in the interval [0,2π) we get 7 values for θ.

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91影视

In Problems 69-78, solve each equation on the interval 0鈮�胃<2蟺.sin2胃+sin4胃=0.

Short Answer

Step by step solution

Step 1Given equation is,

Step 2 Apply the zero product property.

Step 3 Solving the three equations in the interval [0,2π) we get 7 values for θ.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Theoretical and Mathematical Physics

Pure Maths

Geometry

Mechanics Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

In Problems $69 - 78$ , solve each equation on the interval $0 鈮� 胃 < 2 蟺$ .
$\sin (2 胃) + \sin (4 胃) = 0$ .