Methods for finding $\cos(4x)$ given that $\sin(2x) = \frac{3}{5}$ If $\sin(2x) = \frac{3}{5}$
Find $\cos(4x)$..
I tried by : $\cos(4x)= \cos(2\cdot2x)$..
And $\cos (2\cdot2x) = 1-2\sin^2(2x)$ ..
From it ---- $\cos(4x)=0.28$. 
Is there any other ways ? 

Your method seems the 'obvious' way. However you asked for other ways so here is another.
Given $\sin2x=3/5$ you can see from the right angled triangle with opposite $= 3$, hypotenuse $= 5$ that the adjacent side $= 4$ and so $\cos2x=4/5$
Knowing $\sin2x$ and $\cos2x$ you can determine $\sin4x=2\sin2x\cos2x=24/25$
These values are the opposite and hypotenuse of another nice triangle whose adjacent side $=7$. Hence $\cos4x=7/25=0.28$
