Question

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

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

0.4
3.61
1.66
5.81
1.4
3.07

Answer 1

• 3.61
• 5.81

Explanation:

You want solutions to the equation 4sin(θ)² -7sin(θ) -4 = 0 on the interval [0, 2π].

Graph

A graph of the left-side expression shows it has a value of 0 at ...

θ ≈ 3.61 or 5.81.

Check

Using a calculator to check the offered answer choices, we find the values of θ that best satisfy the equation are θ = 3.61 or 5.81. (For this purpose, ±0.02 ≈ 0.)

Algebraic solution

We can let x = sin(θ). Then the equation is ...

4x² -7x -4 = 0

x² -7/4x = 1

(x -7/8)² = 1 +(7/8)² = 113/64

x = (7 ±√113)/8

The value (7+√113)/8 is greater than 1, so the only value of x that is useful here is x = (7-√113)/8. The angles in the desired range are ...

{π - arcsin((7 -√113)/8), 2π +arcsin((7 -√113)/8)} ≈ {3.61, 5.81} . . . radians