find exactly all solutions to the equation?

Question

2cos^2(theta) = 3sin(theta) + 3, for that theta is greater than or eqaul to 0 but less than or equal to 2pi

Answer 1

2 cos ² θ = 3 sin θ + 3
2 (1 - sin ² θ) = 3 sin θ + 3
2 - 2 sin ² θ = 3 sin θ + 3
2 sin ² θ + 3 sin θ + 1 = 0
(2sin θ + 1) (sin θ + 1) = 0
sin θ = - 1/2, sin θ = - 1
θ = 7π/6 , 10π/6 , 3π/2

Answer 2

Let x = theta
2cos^2(x) = 3sin(x) + 3
2(1-sin^2(x) =3sin(x) + 3
2-2sin^2(x) = 3sin(x) +3
2sin^2(x) +3sin(x) +1 =0
sinx = [-3 +/- sqrt(3^2 -4*2*1)]/2*2
sin (x) = [-3 +/- sqrt(1)]/4
sin (x) = -1 and sin(x) = -1/2
Thus x = 210, 270, 330 degrees

Answer 3

2cos2(t) = 3sin(t)+3
2(1-sin2(t)) = 3sin(t) + 3
-1 = 3*(sin(t)) + 2*sin^2(t)
Then use a change of variables,
letting x=sin(t)
then
0= 2x^2 + 3x +1
Use quadratic formula to solve for x
x = (1/4)*(-3 +/- sqrt(9 - 4(2)))
x = (1/4)*(-2 +/- 1)
x1 = -0.25
x2 = -0.75

therefore, as
x1 = sin(t1) = -0.25
t1 = arcsin(-0.25) = -14.77deg = 6.03 rad
t2 = arcsin(-0.75) = -48.6 deg = 5.43 rad