Solve 3-2cos²x-3sinx=0 for 0≤x≤360

3-2 (Cosx) ^2 - 3Sinx = 0.

Recall (Sinx) ^2 + (Cosx) ^2 = 1.

Therefore (Cosx) ^2 = 1 - (Sinx) ^2

Substitute this into the question above.

3-2 (Cosx) ^2 - 3Sinx = 0

3 - 2 (1 - (Sinx) ^2) - 3Sinx = 0 Expand

3 - 2 + 2 (Sinx) ^2 - 3Sinx = 0

1 + 2 (Sinx) ^2 - 3Sinx = 0 Rearrange

2 (Sinx) ^2 - 3Sinx + 1 = 0

Let p = Sinx

2p^2 - 3p + 1 = 0 Factorise the quadratic expression

2p^2 - p - 2p + 1 = 0

p (2p - 1) - 1 (2p - 1) = 0

(2p-1) (p - 1) = 0

Therefore 2p-1=0 or (p-1) = 0

2p=0+1 or (p-1) = 0

2p=1 or p = 0 + 1.

p=1/2 or p = 1 Recall p = Sinx

Therefore Sinx = 1/2 or 1.

For 0
Sinx = 1/2, x = Sin inverse (1/2), x = 30,

(180-30) - 2nd Quadrant = 150 deg

Sinx = 1, x = Sin inverse (1), x = 90

Therefore x = 30,90 & 150 degrees.

Cheers.