How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$? How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$?   
I found the derivative to be $f'(x)= (2\sin 4\theta)(1+\cos 2\theta) + (\sin 2\theta)^2 (-2\sin 2\theta)$.  
Now I know that I have to set this equal to zero and solve for $\theta$ but that is where I get stuck, can I solve for $\theta$ or is there a different way that is easier?

I’d start by rewriting $f(x)$, using the identities $\sin2\theta=2\sin\theta\cos\theta$ and $\cos^2\theta=\frac12(1+\cos2\theta)$:
$$f(x)=\sin^22\theta(1+\cos2\theta)=8\sin^2\theta\cos^4\theta\;.$$
In fact, let’s go a step further and get rid of the sine:
$$f(x)=8\sin^2\theta\cos^4\theta=8(1-\cos^2\theta)\cos^4\theta=8\cos^4\theta-8\cos^6\theta\;.$$
Differentiating and factoring yields
$$f\,'(x)=16\sin\theta\cos^3\theta(3\cos^2\theta-2)\;.$$
Now set this to $0$ and solve. If you want, you can further simplify a little by noting that
$$3\cos^2\theta-2=3(1-\sin^2\theta)-2=1-3\sin^2\theta\;.$$
