Solve for $x$ and represent the solutions on the trig circle? $$\sin(4x)\cos(x)+2\sin(x)\cos^2(2x) = 1- 2\sin^2(x)$$
I'm confused. 

Since $1-2\sin^2(x)=\cos (2x)$ and $\sin (2\alpha)=2\sin (\alpha)\cos (\alpha)$ we have
$$2\sin(2x)\cos(2x)\cos(x)+2\sin(x)\cos^2(2x)=\cos(2x)$$
$$\cos (2x)[2\sin(2x)\cos(x)+2\sin(x)\cos(2x)-1]=0$$
Then either $\cos (2x)=0$ or  $2\sin(2x)\cos(x)+2\sin(x)\cos(2x)-1=0$.
