Hehe tricky equation.

Solve for x:
1-2 cos^2(x)+cos^4(x)+2 sin^2(x)-cos(x) sin(x) sin(2 x) = 0

Reduce trigonometric functions:
-3/8 (-3+4 cos(2 x)-cos(4 x)) = 0

Multiply both sides by -8/3:
-3+4 cos(2 x)-cos(4 x) = 0

Transform -3+4 cos(2 x)-cos(4 x) into a polynomial with respect to cos(2 x) using cos(4 x) = 2 cos^2(2 x)-1:
-2+4 cos(2 x)-2 cos^2(2 x) = 0

Divide both sides by -2:
1-2 cos(2 x)+cos^2(2 x) = 0

Write the left hand side as a square:
(cos(2 x)-1)^2 = 0

Take the square root of both sides:
cos(2 x)-1 = 0

Add 1 to both sides:
cos(2 x) = 1

Take the inverse cosine of both sides:
2 x = 2 π n for n element Z

Divide both sides by 2:
Answer: |x = π n for n element Z

Thats tough

No wonder you gave me a 1 star guys, I made a mistake

The correct equation is: $\cos^4x-(\sin 2x)(\sin x)(\cos x)-2\cos^2x+\sin^4x+2\sin^2x+1=0$

Hi Max :)

\cos^4x-(\sin 2x)(\sin x)(\cos x)-2\cos^2x+\sin^4x+2\sin^2x+1=0

$\cos^4x-(\sin 2x)(\sin x)(\cos x)-2\cos^2x+\sin^4x+2\sin^2x+1=0\\ \cos^4x+\sin^4x-(\sin 2x)(\frac{sin2x}{2})-2(\cos^2x-\sin^2x)+1=0\\ \cos^4x+\sin^4x+2cox^2xsin^2x-2cox^2xsin^2x-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ (\cos^2x+\sin^2x)^2-\frac{4cox^2xsin^2x}{2}-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ 1-\frac{(2cosxsinx)^2}{2}-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ -\frac{(sin2x)^2}{2}-\frac{(sin2x)^2}{2}-2(cos2x)+2=0\\ -(sin2x)^2-2(cos2x)+2=0\\ -[ 1-cos^2(2x) ]-2(cos2x)+2=0\\ -1+(cos2x)^2 -2(cos2x)+2=0\\ (cos2x)^2 -2(cos2x)+1=0\\ let \;\;y=cos2x\\ y^2-2y+1=0\\ (y-1)^2=0\\ y=1\\ cos2x=1\\ 2x=2\pi n \qquad n\in Z\\ x=\pi n \qquad n\in Z\\ $

check

$\cos^4(2\pi n)-(\sin 2*(2\pi n))(\sin (2\pi n))(\cos(2\pi n))-2\cos^2(2\pi n)+\sin^4(2\pi n)+2\sin^2(2\pi n)+1=0\\ \quad 1 \qquad - \qquad 0 \qquad - \qquad 2 \qquad + \qquad 0 \qquad + \qquad 0 \qquad + \qquad 1 \qquad = \qquad 0 \qquad GOOD$