Did you mean cos(2⁡θ) or cos^2(⁡θ)?

I'll assume the latter, since the former has no real solutions.

But you should never write cos^2(⁡θ) as cos2⁡θ.

2 cos²θ = 3 sinθ + 3

2 (1 − sin²θ) = 3 sinθ + 3

2 − 2 sin²θ = 3 sinθ + 3

2 sin²θ + 3 sinθ + 1= 0

(sinθ + 1) (2 sinθ + 1) = 0

sinθ = −1 ----> θ = 3π/2

sinθ = −1/2 ----> θ = 7π/6, 11π/6

Answers in increasing order: 7π/6, 3π/2, 11π/6

There are only 3 answers, not 4

cos²(θ) = 1 - sin²(θ)

2 cos²(θ) = 3 sin(θ) + 3

2 (1 - sin²(θ)) = 3 sin(θ) + 3

2 - 2sin²(θ) = 3 sin(θ) + 3

0 = 2sin²(θ) + 3 sin(θ) + 1

0 = (2sin(θ) + 1) (sin(θ) + 1)

2sin(θ) + 1 = 0

2sin(θ) = -1

sin(θ) = -1/2

θ = 7π/6, 11π/6

sin(θ) + 1 = 0

sin(θ) = -1

θ = 3π/2

θ = 7π/6, 3π/2, 11π/6

There are only three solutions in the interval 0≤θ≤2π.