Precalc math problem

Question

Solve the equation for solutions over the interval

​[0°​,360°​).

6cos²θsinθ−3cos²θ=0

Answer 1

6cos^2(T)sinT -3cos^2(T) = 0

3cos^2(T)[2sinT -1) = 0

cos^2(T) = 0

T = 90 or 270 degrees

sinT = 1/2

T = 30 or 150 degrees

4 solutions:

T = 30, 90, 150 or 270 degrees