Which shows all the exact solutions of sin^2 x + 3cos x - 1 = 2

Answer: x = π/2 + 2πn, for n ∈ N

Explanation:

1) Given: sin²x + 3 cosx - 1 = 2

2) Make sin²x = 1 - cos²x ⇒ 1 - cos²x + 3cosx - 1 = 2

3) Add like terms and transpose terms to equal 0:

cos²x - 3cosx + 2 = 0

4) factor: (cos x - 2) (cosx - 1) = 0

5) Find the solution for each factor:

cosx - 2 = 0⇒ cosx = 2 ↔ impossible as cosx is between - 1 and 1.

cos x - 1 = 0 ⇒ cosx = 1 ⇒ x = π/2 + 2πn, for n ∈ N