Approximate the first three non-negative solutions to: 3sin2(x)−2cos(x)=2

Question

Approximate the first three non-negative solutions to:

3sin2(x)−2cos(x)=2

Give your answers in degrees to one decimal place.

Answer 1

3sin²(x)-2cox(x)=2

3(1-cos²(x))-2cos(x)-2=0

3-3cos²(x)-2cos(x)-2 = 0

3cos²(x)-2cos(x)-1=0

(cos(x)+1)(3cos(x)-1) =0

cos(x) = -1 --> x = 180°

cos(x)=1/3 --> cos^-1(1/3) = 70.5°,360-70.5 = 289.5°

x = 70.5°, 180°, 289.5°