Question

Solve the following for θ, in radians, where 0≤θ<2π.

4sin2(θ)+7sin(θ)−5=0
Select all that apply:

2.57
1.37
1.05
2.35
0.58
1.88

Answer 1

Answer:θ ≈ 2.57 radians

θ ≈ 0.58 radians

Step-by-step explanation:We can solve this quadratic equation in sin(θ) by factoring:

4sin^2(θ) + 7sin(θ) - 5 = 0

(4sin(θ) - 1)(sin(θ) + 5) = 0

Therefore, either:

4sin(θ) - 1 = 0

sin(θ) = 1/4

or:

sin(θ) + 5 = 0

sin(θ) = -5 (not possible, since sin(θ) is between -1 and 1)

So, we have sin(θ) = 1/4. Since 0 ≤ θ < 2π, we can find the two solutions in the interval [0, 2π) by using the inverse sine function:

θ = arcsin(1/4)

Using a calculator, we find:

θ ≈ 0.2531 or θ ≈ 2.8887

Therefore, in radians, the solutions for θ are approximately:

0.2531 (which is less than 2π)

2.8887 (which is greater than 2π)

So the only answer that satisfies 0 ≤ θ < 2π is: