Solve 2 + 2 sin θ = 3 cos^2
θ. Give the general solution and then give solutions in [−π, π]
How do i solve for theta?
2 + 2 sinθ = 3 cos^2θ
2 + 2sinθ = 3 - 3sin^2θ
3sin^2θ + 2sinθ - 1 = 0
(3sinθ-1)(sinθ+1) = 0
sinθ = 1/3
sinθ = -1
so, there are two solutions for sinθ = 1/3 (QI and QII)
and one solution for sinθ = -1