Chapter 7: Q. 77 (page 496)

In Problems 69-78, solve each equation on the interval $\tan (2 θ) + 2 \sin θ = 0$

Short Answer

Expert verified

The solution set is $θ \in \{\frac{2 π}{3}, \frac{4 π}{3}\}$

Step by step solution

Given information

Given function $\tan (2 θ) + 2 \sin θ = 0$

Using the double angle formula and factoring

Using the formula, we get

$\cos (2 θ) + 5 \cos θ + 3 = 0 2 \cos^{2} θ - 1 + 5 \cos θ + 3 = 0 2 \cos^{2} θ + 5 \cos θ + 2 = 0 (2 \cos θ + 1) (\cos θ + 2) = 0 2 \cos θ + 1 = 0 \cos θ = - \frac{1}{2} \cos θ + 2 = 0 \cos θ = - 2 θ = \frac{2 π}{3} θ = \frac{4 π}{3}$

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In Problems 69-78, solve each equation on the interval $\tan (2 θ) + 2 \sin θ = 0$

Short Answer

Step by step solution

Given information

Using the double angle formula and factoring

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In Problems 69-78, solve each equation on the intervaltan(2θ)+2sinθ=0

Short Answer

Step by step solution

Given information

Using the double angle formula and factoring

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Discrete Mathematics

Mechanics Maths

Theoretical and Mathematical Physics

Applied Mathematics

Statistics

Study anywhere. Anytime. Across all devices.

In Problems 69-78, solve each equation on the interval $\tan (2 θ) + 2 \sin θ = 0$