Chapter 7: Q. 69 (page 496)

In Problems 69–78, solve each equation on the interval $0 \leq θ < 2 π$ .
$\cos (2 θ) + 6 \sin^{2} θ = 4$ .

Short Answer

Expert verified

The solution set for the equation $\cos (2 θ) + 6 \sin^{2} θ = 4$ is, $\{\frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}\}$ .

Step by step solution

Step 1 Given equation is,

$\cos (2 θ) + 6 \sin^{2} θ = 4$ .

Th double angle formula for $\cos (2 θ)$ is, $\cos (2 θ) = 1 - 2 \sin^{2} θ$ .

Use the formula in the given equation.

$1 - 2 \sin^{2} θ + 6 \sin^{2} θ = 4$

Now simplify it.

$- 2 \sin^{2} θ + 6 \sin^{2} θ = 4 - 1 4 \sin^{2} θ = 3 \sin^{2} θ = \frac{3}{4}$

Step 2 Now take the square root on both the sides of the equation.

$\sin θ = \pm \sqrt{\frac{3}{4}} \sin θ = \pm \frac{\sqrt{3}}{2}$

Solving each equation in the interval $[0, 2 π)$ we obtain $θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}$ and $θ = \frac{5 π}{3}$ .

Therefore the solution set is, $\{\frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}\}$ .

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In Problems 69–78, solve each equation on the interval $0 \leq θ < 2 π$ .
$\cos (2 θ) + 6 \sin^{2} θ = 4$ .

Short Answer

Step by step solution

Step 1 Given equation is,

Step 2 Now take the square root on both the sides of the equation.

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In Problems 69–78, solve each equation on the interval 0≤θ<2π.cos2θ+6sin2θ=4.

Short Answer

Step by step solution

Step 1 Given equation is,

Step 2 Now take the square root on both the sides of the equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Mechanics Maths

Calculus

Statistics

Logic and Functions

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Study anywhere. Anytime. Across all devices.

In Problems 69–78, solve each equation on the interval $0 \leq θ < 2 π$ .
$\cos (2 θ) + 6 \sin^{2} θ = 4$ .