Chapter 7: Q. 15 (page 465)

In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$ .
$2 \sin^{2} θ - 1 = 0$

Short Answer

Expert verified

The solution set is $\{\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}\}$ .

Step by step solution

Step 1. Given Information

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

$2 \sin^{2} θ - 1 = 0$

Step 2. Firstly solving the equation 2sin2θ-1=0

Add 1 on both side

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

Divide by 2 on both side

$\frac{2}{2} \sin^{2} θ = \frac{1}{2} \sin^{2} θ = \frac{1}{2}$

Taking square root on both side

$\sin θ = \pm \sqrt{\frac{1}{2}} \sin θ = \pm \frac{1}{\sqrt{2}}$

$\sin θ = \frac{1}{\sqrt{2}} and \sin θ = - \frac{1}{\sqrt{2}}$

Step 3. After solving the equation sinθ=12 and sinθ=-12.

In the interval $[0, 2 π)$ , there are two angles $θ$ for which $\sin θ = \frac{1}{\sqrt{2}} : θ = \frac{π}{4} and θ = \frac{3 π}{4}$

Step 4. The period of the cosine function sinθ=-12 is 2π.

In the interval $[0, 2 π)$ , there are two angles $θ$ for which $\sin θ = - \frac{1}{\sqrt{2}} : θ = \frac{5 π}{4} and θ = \frac{7 π}{4}$

The solution set is $\{\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}\}$ .

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In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$ .
$2 \sin^{2} θ - 1 = 0$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Firstly solving the equation 2sin2θ-1=0

Step 3. After solving the equation sinθ=12 and sinθ=-12.

Step 4. The period of the cosine function sinθ=-12 is 2π.

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In Problems 11–34, solve each equation on the interval 0≤θ≤2π.2sin2θ-1=0

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Firstly solving the equation 2sin2θ-1=0

Step 3. After solving the equation sinθ=12 and sinθ=-12.

Step 4. The period of the cosine function sinθ=-12 is 2π.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Applied Mathematics

Geometry

Theoretical and Mathematical Physics

Mechanics Maths

Decision Maths

Study anywhere. Anytime. Across all devices.

In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$ .
$2 \sin^{2} θ - 1 = 0$