Q. 71 In Problems 57鈥� 80, solve each... [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. 71 (page 465)

In Problems 57鈥� 80, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$
localid="1647006656862" $1 + \sin 胃 = 2 \cos^{2} 胃$

Short Answer

Expert verified

The solution set of the given equation is $\{\frac{蟺}{6}, \frac{5 蟺}{6}, \frac{3 蟺}{2}\}$ .

Step by step solution

Step 1. Given Information

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

$1 + \sin 胃 = 2 \cos^{2} 胃$

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

However, a form of the Pythagorean Identity, $\sin^{2} 胃 + \cos^{2} 胃 = 1$ , can be used to transform the equation into an equivalent one containing only sines.

1 + \sin 胃 = 2 (1 - \sin^{2} 胃) 1 + \sin 胃 = 2 - 2 \sin^{2} 胃

Subtract 2 and add $2 \sin^{2} 胃$ on both side

$1 + \sin 胃 - 2 + 2 \sin^{2} 胃 = 2 - 2 \sin^{2} 胃 - 2 + 2 \sin^{2} 胃 \sin 胃 - 1 + 2 \sin^{2} 胃 = 0 2 \sin^{2} 胃 + \sin 胃 - 1 = 0$

Step 3. Now the equation is 2sin2θ+sinθ-1=0.

Factor the equation.

$2 \sin^{2} 胃 + (2 - 1) \sin 胃 - 1 = 0 2 \sin^{2} 胃 + 2 \sin 胃 - \sin 胃 - 1 = 0 2 \sin 胃 (\sin 胃 + 1) - 1 (\sin 胃 + 1) = 0 (2 \sin 胃 - 1) (\sin 胃 + 1) = 0$

Step 4. Use the Zero-Product Property.

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

Step 5. Solving each equation in the interval [0,2π], we obtain

$胃 = \frac{蟺}{6}, \frac{5 蟺}{6} 胃 = \frac{3 蟺}{2}$

The solution set is $\{\frac{蟺}{6}, \frac{5 蟺}{6}, \frac{3 蟺}{2}\}$

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91影视

In Problems 57鈥� 80, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$
localid="1647006656862" $1 + \sin 胃 = 2 \cos^{2} 胃$

Short Answer

Step by step solution

Step 1. Given Information

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

Step 3. Now the equation is 2sin2θ+sinθ-1=0.

Step 4. Use the Zero-Product Property.

Step 5. Solving each equation in the interval [0,2π], we obtain

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91影视

In Problems 57鈥� 80, solve each equation on the interval 0鈮�胃鈮�2蟺localid="1647006656862" 1+sin胃=2cos2胃

Short Answer

Step by step solution

Step 1. Given Information

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

Step 3. Now the equation is 2sin2θ+sinθ-1=0.

Step 4. Use the Zero-Product Property.

Step 5. Solving each equation in the interval [0,2π], we obtain

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Decision Maths

Applied Mathematics

Discrete Mathematics

Probability and Statistics

Pure Maths

Study anywhere. Anytime. Across all devices.

In Problems 57鈥� 80, solve each equation on the interval $0 鈮� 胃鈮� 2 蟺$
localid="1647006656862" $1 + \sin 胃 = 2 \cos^{2} 胃$