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. 

continue @Mario G'answer with $2\sin(2x)\cos(x)+2\sin(x)\cos(2x)-1=0 \Rightarrow‎ 2sin3x-1=0\Rightarrow‎ sin3x=\frac{1}{2}$.....
$\sin(a+b)=\sin a \cos b+\sin b \cos a$
