Chapter 7: Q. 69 (page 465)

In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
$\tan θ = 2 \sin θ$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information

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

$\tan θ = 2 \sin θ$

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

We know that $\tan θ = \frac{\sin θ}{\cos θ}$

So $\frac{\sin θ}{\cos θ} = 2 \sin θ$

Divide by $\sin θ$ on both side

$\frac{\sin θ}{\cos θ} \cdot \frac{1}{\sin θ} = 2 \frac{\sin θ}{\sin θ} \frac{1}{\cos θ} = 2$

Step 3. We know that secθ=1cosθ

So $\sec θ = 2$

Solving each equation in the interval $[0, 2 π]$ , we obtain

$θ = \frac{π}{3}, \frac{5 π}{3}$

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

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In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
$\tan θ = 2 \sin θ$

Short Answer

Step by step solution

Step 1. Given Information

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

Step 3. We know that secθ=1cosθ

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In Problems 57– 80, solve each equation on the interval 0≤θ≤2πtanθ=2sinθ

Short Answer

Step by step solution

Step 1. Given Information

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

Step 3. We know that secθ=1cosθ

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Statistics

Mechanics Maths

Probability and Statistics

Applied Mathematics

Calculus

Study anywhere. Anytime. Across all devices.

In Problems 57– 80, solve each equation on the interval $0 \leq θ \leq 2 π$
$\tan θ = 2 \sin θ$