Chapter 7: Q. 21 (page 465)

In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$ .
$s e c \frac{3 θ}{2} = - 2$

Short Answer

Expert verified

The solution set is $\{\frac{4 π}{9}, \frac{8 π}{9}, \frac{16 π}{9}\}$ .

Step by step solution

Step 1. Given Information

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

$s e c \frac{3 θ}{2} = - 2$

Step 2. In the interval [0,2π), the sine function equals -2 at 2π3

So, we know that $\frac{3 θ}{2}$ must equal $\frac{2 π}{3}$ .

To find these solutions, write the general formula that gives all the solutions.

role="math" localid="1646590784861" $\frac{3 θ}{2} = \frac{2 π}{3} + 2 πn$

Multiply with $\frac{2}{3}$ on both side

role="math" localid="1646591123910" $\frac{3 θ}{2} \cdot \frac{2}{3} = \frac{2}{3} (\frac{2 π}{3} + 2 πn) θ = \frac{2}{3} \cdot \frac{2 π}{3} + \frac{2}{3} \cdot 2 πn θ = \frac{4 π}{9} + \frac{4 πn}{3}$

$θ = \frac{8 π}{9} + \frac{4 πn}{3}$

Step 3. The general formula is θ=4π9+4πn3,θ=8π9+4πn3

So the value of given function in interval $[0, 2 π)$ is

role="math" localid="1646591235840" $θ = \frac{4 π}{9} + \frac{4 π \times 0}{3} θ = \frac{4 π}{9} + \frac{4 π \times 1}{3} θ = \frac{8 π}{9} + \frac{4 π \times 0}{3} θ = \frac{4 π}{9} θ = \frac{4 π}{9} + \frac{4 π}{3} θ = \frac{8 π}{9} θ = \frac{4 π}{9} θ = \frac{4 π + 12 π}{9} θ = \frac{4 π}{9} θ = \frac{16 π}{9}$

So the solution set is $\{\frac{4 π}{9}, \frac{8 π}{9}, \frac{16 π}{9}\}$

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In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$ .
$s e c \frac{3 θ}{2} = - 2$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. In the interval [0,2π), the sine function equals -2 at 2π3

Step 3. The general formula is θ=4π9+4πn3,θ=8π9+4πn3

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In Problems 11–34, solve each equation on the interval 0≤θ≤2π.sec3θ2=-2

Short Answer

Step by step solution

Step 1. Given Information

Step 2. In the interval [0,2π), the sine function equals -2 at 2π3

Step 3. The general formula is θ=4π9+4πn3,θ=8π9+4πn3

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Decision Maths

Discrete Mathematics

Theoretical and Mathematical Physics

Geometry

Pure Maths

Study anywhere. Anytime. Across all devices.

In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$ .
$s e c \frac{3 θ}{2} = - 2$