Chapter 7: Q. 70 (page 496)

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

Short Answer

Expert verified

The equation $\cos (2 θ) = 2 - 2 \sin^{2} θ$ has no solution (i.e), $θ \in \{ϕ\}$ .

Step by step solution

Step 1 Given equation is,

$\cos (2 θ) = 2 - 2 \sin^{2} θ$ .

we know that, $\cos (2 θ) = 1 - 2 \sin^{2} θ$

Substitute the known values in the given equation.

$1 - 2 \sin^{2} θ = 2 - 2 \sin^{2} θ 1 \neq 2$

Step 2 The above equation indicates that the given equation has no solution.

So, the solution set is, $θ \in {ϕ}$ .

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

Short Answer

Step by step solution

Step 1 Given equation is,

Step 2 The above equation indicates that the given equation has no solution.

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

Short Answer

Step by step solution

Step 1 Given equation is,

Step 2 The above equation indicates that the given equation has no solution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Statistics

Pure Maths

Theoretical and Mathematical Physics

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Calculus

Study anywhere. Anytime. Across all devices.

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