sqrt(4 - (2sinx)^2)2cosx = 4cos^2(x)
There are steps in the middle (obviously) but I can't figure them out
I can see why you are having trouble: the statement is not an identity. For example, with $\displaystyle x=\pi,$ the left-hand side is $\displaystyle 2(-2)=-4,$ but the right-hand side is $\displaystyle 4(-1)^2=4.$ We can, however, show that the two expressions differ only in sign:
$\displaystyle 2\cos x\sqrt{4-(2\sin x)^2}$
$\displaystyle =2\cos x\sqrt{4-4\sin^2x}$
$\displaystyle =2\cos x\sqrt{4\left(1-\sin^2x\right)}$
$\displaystyle =4\cos x\sqrt{1-\sin^2x}$
$\displaystyle =4\cos x\sqrt{\cos^2x}$
$\displaystyle =4\cos x|\cos x|$
$\displaystyle =\pm4\cos^2x$