Chapter 7: Q. 20 (page 465)

In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$
$\tan (2 θ) = - 1$

Short Answer

Expert verified

The solution set is $\{\frac{3 π}{8}, \frac{7 π}{8}, \frac{11 π}{8}, \frac{15 π}{8}\}$ .

Step by step solution

Step 1. Given Information

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

$\tan (2 θ) = - 1$

Step 2. In the interval [0,2π), the tangent function equals -1 at 3π4

$2 θ$ must equal to $\frac{3 π}{4}$ .

$2 θ = \frac{3 π}{4} + π n$

Divide by 2 on both side

$\frac{2}{2} θ = \frac{1}{2} (\frac{3 π}{4} + π n) θ = \frac{1}{2} \cdot \frac{3 π}{4} + \frac{1}{2} \cdot π n θ = \frac{3 π}{8} + \frac{π n}{2}$

Step 3. The general formula is θ=3π8+πn2

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

$θ = \frac{3 π}{8} + \frac{π \times 0}{2} θ = \frac{3 π}{8} + \frac{π \times 1}{2} θ = \frac{3 π}{8} + \frac{π \times 2}{2} θ = \frac{3 π}{8} + \frac{π \times 3}{2} θ = \frac{3 π}{8} θ = \frac{3 π}{8} + \frac{π}{2} θ = \frac{3 π}{8} + π θ = \frac{3 π}{8} + \frac{3 π}{2} θ = \frac{3 π}{8} θ = \frac{3 π + 4 π}{8} θ = \frac{3 π + 8 π}{8} θ = \frac{3 π + 12 π}{8} θ = \frac{3 π}{8} θ = \frac{7 π}{8} θ = \frac{11 π}{8} θ = \frac{15 π}{8}$

So the solution set is $\{\frac{3 π}{8}, \frac{7 π}{8}, \frac{11 π}{8}, \frac{15 π}{8}\}$ .

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In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$
$\tan (2 θ) = - 1$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. In the interval [0,2π), the tangent function equals -1 at 3π4

Step 3. The general formula is θ=3π8+πn2

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In Problems 11–34, solve each equation on the interval 0≤θ≤2πtan(2θ)=-1

Short Answer

Step by step solution

Step 1. Given Information

Step 2. In the interval [0,2π), the tangent function equals -1 at 3π4

Step 3. The general formula is θ=3π8+πn2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Statistics

Decision Maths

Logic and Functions

Discrete Mathematics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 11–34, solve each equation on the interval $0 \leq θ \leq 2 π$
$\tan (2 θ) = - 1$