Please help me and answer!

Question

Find all angles, 0≤θ360, that satisfy the equation below, to the nearest 10th of a degree.

3 sin² θ+ 3sin θ-6= -4sin θ

Answer 1

3sin²(θ) + 3sinθ) - 6 = -4sin(θ)

3sin²(θ) + 3sinθ) + 4sin(θ) - 6 = 0

3sin²(θ) + 7sinθ) - 6 = 0

Let x = sin(θ) then 3sin²(θ) + 7sinθ) - 6 = 0 becomes:

3x² + 7x - 6 = 0

(x + 3)(3x - 2) = 0

x = -3

x = 2/3

Back substitute x = sin(θ) to get sin(θ) = -3 and sin(θ) = 2/3

sin(θ) is always between -1 and 1 so there are no values of θ for sin(θ) to be -3 so we can throw that out.

For sin(θ) = 2/3, θ = sin^-1(2/3). Using a calculator, we see that θ = 41.8°. But there is a quadrant 2 angle that also complies. It would be 180 - 41.8° so 138.2°

So θ = 41.8° and 138.2°