Chapter 7: Q. 74 (page 504)

Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$\sin (2 θ) = \sqrt{2} \cos θ$

Short Answer

Expert verified

On solving the equation, we get,

$θ \in \{\frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}\}$

Step by step solution

Step 1. Given information.

Consider the given question,

$\sin (2 θ) = \sqrt{2} \cos θ 0 \leq θ \leq 2 π$

We know $\sin (2 θ) = 2 \sin θ \cdot \cos θ$ ,

$2 \sin θ \cdot \cos θ = \sqrt{2} \cos θ 2 \sin θ \cdot \cos θ - \sqrt{2} \cos θ = 0 c os θ (2 \sin θ - \sqrt{2}) = 0$

Then,

localid="1646750340707" $\cos θ = 0$

Also,

$2 \sin θ - \sqrt{2} = 0 \sin θ = \frac{\sqrt{2}}{2} \sin θ = \frac{1}{\sqrt{2}}$

Step 2. Solve the equation for the given interval.

Consider the given question,

$\cos θ = 0 \sin θ = \frac{1}{\sqrt{2}} 0 \leq θ \leq 2 π$

Solving $\cos θ = 0$ for the given interval,

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

Solving role="math" localid="1646750410065" $\sin θ = \frac{1}{\sqrt{2}}$ for the given interval,

$\sin θ = \frac{1}{\sqrt{2}} θ = \sin^{- 1} (\frac{1}{\sqrt{2}}) θ = \frac{π}{4}, \frac{3 π}{4}$

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

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Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$\sin (2 θ) = \sqrt{2} \cos θ$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Solve the equation for the given interval.

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Solve the equation on the interval 0≤θ≤2π.sin2θ=2cosθ

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Solve the equation for the given interval.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Study anywhere. Anytime. Across all devices.

Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$\sin (2 θ) = \sqrt{2} \cos θ$