Chapter 7: Q. 71 (page 465)

In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
localid="1647006656862" $1 + \sin θ = 2 \cos^{2} θ$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information

In the given problems we have to solve each equation on the interval $0 \leq θ \leq 2 π$

$1 + \sin θ = 2 \cos^{2} θ$

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

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

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

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

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

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

Factor the equation.

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

Step 4. Use the Zero-Product Property.

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

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

$θ = \frac{π}{6}, \frac{5 π}{6} θ = \frac{3 π}{2}$

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

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In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
localid="1647006656862" $1 + \sin θ = 2 \cos^{2} θ$

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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In Problems 57– 80, solve each equation on the interval 0≤θ≤2πlocalid="1647006656862" 1+sinθ=2cos2θ

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

Probability and Statistics

Calculus

Theoretical and Mathematical Physics

Discrete Mathematics

Logic and Functions

Statistics

Study anywhere. Anytime. Across all devices.

In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
localid="1647006656862" $1 + \sin θ = 2 \cos^{2} θ$