Solve the equation for the prinicipal values of x 2sin2x 3c

Solve the equation for the prinicipal values of x:

2sin^2x + 3cosx -3 = 0

2sin^2 x +3cosx -3=0

but sin^2 x +cos^2 x =1

so sin^2 x = 1-cos^2 x

2[ 1-cos^2 x ] +3cosx -3=0

2 -2cos^2x -3cosx -3=0

-2cos^2 x -3cosx -1=0

-[2cos^2 x +3cosx+1]=0

2cos^2 x +3cosx+1 =0

2cos^2 x +2cosx +cosx+1 =0

2cosx(cosx +1) +1(cosx+1) =0

(cosx+1) (2cosx+1) =0

cosx+1 =0 or 2cosx+1=0

cosx =-1 cosx =-1/2

x= npi n =1,3 ,5 ,,2n-1

x = npi+pi/3