Find, to the nearest degree, all values of x in the the interval [0 degrees, 360 degrees) that satisfy the equ

Question

Find, to the nearest degree, all values of x in the the interval [0 degrees, 360 degrees) that satisfy the equation

3 cos 2x + sin x - 1 = 0.

I know the answer, i just can't figure out how to get it.

Answer 1

cos(2x) = cos^2x-sin^2x= 1-2sin^2 x
So 3 -6 sin^2x +sinx -1=0

6sin^2x -sinx-2=0 put sinx=z

6z^2 -z -2 =0 2nd degree equation

z= ((1+-sqrt(1+48))/12 = (1+- 7)/12

z=8/12=2/3 and z= -1/2

sin x=2/3 x= 41.8 degrees and 138.19 degrees

sin x= -1/2 x=210 deg and x= 330 deg.