<PRE><B>Question 901554</B>
sin^2(2x)+sin^4x=2.

(1-cos(4x))/2 +(1-cos^2x)^2=0

(1-cos(4x))/2 +(1-2cos^2x+cos^4x)=0

(1-(8cos^4(x) - 8cos^2(x) + 1 )/2+1-2cos^2+cos^4x=0

1-8cos^4x+8cos^2x-1+2-4cos^2x+2cos^4x=0

-6cos^4x+4cos^2x+2=0

6cos^4x-4cos^2x-2=0

6cos^4x-6cos^2x+2cos^2x-2=0

6cos^2x(cos^2x-1)+2(cos^2-1)=0

(cos^2x-1)(6cos^2x+2)=0

(cos^2x-1)(3cos^2x+1)=0



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