Solving for x in trigonometry equations

I need to find the values for x in the trigonometric equation(correct to one decimal digit)
2sin2x - 2sinx + 3cosx = 6cos(^2)x
If possible please can you show all of the steps
thank you in advance

2.2sinx.cosx -2sinx +3cosx=6co(^2)x

4sinxcosx- 2sinx -6cos^2x+3cosx=0 2sinx(2cosx-1)-3cosx(2cosx-1)=0 (2cosx-1)( 2sinx-3cosx)=0

2cosx-1=0 or 2sinx-3cosx=0 cosx=1/2 or tanx=3/2 x=pi/3 x=tan^-1(3/2)

2sin2x - 2sinx + 3cosx = 6cos(^2)x
OR 2(2 sin x cos x) -- 2 sin x = 6 cos^2 x -- 3 cos x
OR 2 sin x [2 cos x -- 1] = 3 cos x [ 2 cos x -- 1]
OR (2 cos x -- 1)(2 sin x -- 3 cos x) = 0
Giving cos x = 1/2 = cos (pi/3) whence x = 2n(pi) +/-- (pi/3)
tan x = 3/2 whence x = n(pi) + arctan (3/2)