Given that $\piThe left side of the expression reduces to $2\sin x$ and the right side reduces $2+2\sin x$ 
Their addition gives $2+4\sin x$ . The correct answer is 2
I realize that this has got to do with the fact that $\pi<x<3\pi/2$ , in which case $\sin x$ will be -ve. 
But then the expression should be $2-4\sin x$.  How does it end up being 2?

$$\sqrt{4\sin^4x+\sin^2(2x)}+4\cos^2\left(\frac\pi4-\frac x2\right)$$
$$=\sqrt{4\sin^4x+4\sin^2x\cdot\cos^2x}+2\left(\cos\left(\frac\pi2-x\right)+1\right)$$
$$=2\vert \sin x \vert \sqrt{\sin^2x+\cos^2x}+2\sin x+2$$
$$=-2\sin x+2\sin x+2=2$$
