Can someone help me with this trigonometry/calculus proof?

Question

4(cos^6x+sin^6x) = 1+3cos^2 2x

prove that 4(cos^6 x + sin^6 x) is equal to 1+3cos^2 2x

Answer 1

4(cos^6x+sin^6x) = 4 (cos^2x + sin^2x)(sin^4x + cos^4x - sin^2x*cos^2x)
= 4 ( 1 - 3sin^2x*cos^2x) = 4 - 6 sin2x
=>1 + 3cos^2 2x = 4 - 6sin2x
<=>4 - 3sin^2 2x = 4 - 6 sin2x
<=> sin^2 2x - 2sin2x = 0
<=> sin2x( sin2x - 2) =0
<=> sin2x =0 <=> x = k*pi/2 ( k * 90 ) with k is an integer