Chapter 7: Q. 77 (page 496)

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

Short Answer

Expert verified

The solution set is $\{0, \frac{π}{3}, π, \frac{5 π}{3}\}$

Step by step solution

Given information

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

rewrite the expression in terms of sine and cosine

Rewriting, we get

$\tan θ = \frac{\sin θ}{\cos θ} \frac{\sin 2 θ}{\cos 2 θ} + 2 \sin θ = 0 \sin 2 θ + 2 \sin θ \cos 2 θ = 0 2 \sin θ \cos θ + 2 \sin θ \cos 2 θ = 0 2 \sin θ (\cos θ + \cos 2 θ) = 0$

further solving using identities

Further solving, we get

$2 \sin θ = 0 \sin θ = 0 θ = 0, π$

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

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

Short Answer

Step by step solution

Given information

rewrite the expression in terms of sine and cosine

further solving using identities

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In Problems 69-78, solve each equation on the interval 0≤θ<2πtan(2θ)+2sinθ=0

Short Answer

Step by step solution

Given information

rewrite the expression in terms of sine and cosine

further solving using identities

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

Mechanics Maths

Logic and Functions

Decision Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

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