Question

sin^4x+cos^4x=cos^2 2 x

Answer 1

$sin^4x+cos^4x=cos^22x\\sin^4x+cos^4x=(cos^2x-sin^2x)^2\\star [, cos2x=cos^2x-sin^2x, ]star \\sin^4x+cos^4x=cos^4x-2sin^2xcdot cos^2x+sin^4x\\0=-2sin^2xcdot cos^2x\\sin^2xcdot cos^2x=0\\star [, 2sinxcdot cosx=sin2x, ,; 4sin^2xcdot cos^2x=sin^22x]star \\frac{sin^22x}{4}=0\\sin^22x=0; ; Rightarrow sin2x=0\\2x=pi n ,; nin Z\\x=frac{pi n}{2}; ,; nin Z$