Chapter 7: Q. 73 (page 504)

Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$4 \sin^{2} θ = 1 + 4 \cos θ$

Short Answer

Expert verified

On solving the equation, we get,

$θ \in \{\frac{π}{3}, \frac{5 π}{3}\}$

Step by step solution

Step 1. Given information.

Consider the given question,

$4 \sin^{2} θ = 1 + 4 \cos θ 0 \leq θ \leq 2 π$

We know, $\sin^{2} θ = 1 - \cos^{2} θ$ ,

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

Then,

localid="1646749506085" $2 \cos θ - 1 = 0 \cos θ = \frac{1}{2}$

Also,

$2 \cos θ + 3 = 0 \cos θ = - \frac{3}{2}$

This equation is invalid as cosine value must lie within the range $[- 1, 1]$ .

Step 2. Solve the equation for the given interval.

Consider the given question,

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

Solving $\cos θ = \frac{1}{2}$ for the given interval,

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

Therefore, the solution set is $θ \in \{\frac{π}{3}, \frac{5 π}{3}\}$ .

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Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$4 \sin^{2} θ = 1 + 4 \cos θ$

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.4sin2θ=1+4cosθ

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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Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Solve the equation on the interval $0 \leq θ \leq 2 π$ .
$4 \sin^{2} θ = 1 + 4 \cos θ$