show that 2cos(2x) - cos^2(x) = 1 -3sin^2(x)

Verify the following identity:
2 cos(2 x)-cos(x)^2  =  1-3 sin(x)^2

cos(x)^2 = 1-sin(x)^2:
2 cos(2 x)-1-sin(x)^2  =  ^?1-3 sin(x)^2

cos(2 x) = 1-2 sin(x)^2:
2 1-2 sin(x)^2-(1-sin(x)^2)  =  ^?1-3 sin(x)^2

-(1-sin(x)^2) = sin(x)^2-1:
sin(x)^2-1+2 (1-2 sin(x)^2)  =  ^?1-3 sin(x)^2

2 (1-2 sin(x)^2) = 2-4 sin(x)^2:
-1+sin(x)^2+2-4 sin(x)^2  =  ^?1-3 sin(x)^2

-1+sin(x)^2+2-4 sin(x)^2 = 1-3 sin(x)^2:
1-3 sin(x)^2  =  ^?1-3 sin(x)^2

The left hand side and right hand side are identical:
Answer: | identity has been verified)
 

Here's another method :

2cos(2x) - cos^2(x) = 1 -3sin^2(x)          break up  -3sin^2(x)  into  -2sin^2x - sin^2(x) 

2cos(2x) - cos^2(x)  = 1 - 2sin^2(x) -sin^2(x)     rearrange the right side

2cos(2x) - cos^2x  = 1 - sin^2(x) - 2sin^2(x)

2cos(2x) - cos^2(x)  = cos^2(x) - 2sin^2(x)    add cos^2(x) to both sides

2cos(2x)  = 2cos^2(x) - 2sin^2(x)    divide through by 2

cos(2x)  = cos^2(x) - sin^2(x)         which is a basic  identity