Solve the equation exactly over the interval [0,360°). sin^2θ - sin θ - 12 = 0?

Answer 1

Notice that this is a quadratic in sin x (I'll use x instead of theta, as x is easier to type). So factor (or use quadratic formula):
sin^2(x) - sin(x) - 12 = 0
(sin(x) - 4) (sin(x) + 3) = 0
sin(x)= 4 or sin(x)= -3.

Well, here we have a problem. The sine of any angle must have a value between -1 and 1 (unless we consider imaginary values). Either there's a problem in this question, or there is no solution. Perhaps you meant "-2" instead of 12:
(sin(x) + 1) (sin(x) - 2) = 0
sin(x)= -1 or sin(x)= 2.
Here only the first option is valid, so sin(x)= -1, so that x=270 degrees.

Answer 2

sin ² θ - sin θ - 12 = 0
(sinθ - 4) (sin θ + 3) = 0
sin θ = 4 , sin θ = - 3
Error in question.