<s>[INST] What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \sin x \sin {2x}$$
I have done my work here 
$$f (x)=\sin x \sin 2x =\frac{\cos x - \cos3x}2 $$
$$f'(x)= \frac{- \sin x+3 \sin 3x}2 =4\sin x (2-3\sin^2 x)=0$$
$$x=0,\pi; \sin x= \sqrt{\frac {2}{3}}$$
$$f (0)=f (\pi)=0$$
$$f \left(\arcsin \sqrt{\frac{2}{3}}\right) =\frac{4}{3 \sqrt{3}}$$
If my work is not much right then please rectify it
 [/INST] This is a slight simplification of Seyed's solution.
We have $$f(x) = \sin (x)\sin (2x) = 2(\cos (x) - {\cos ^3}(x))=2g(\cos(x))$$
for an extreme we should have 
$$0=g'(y) = 1-3 y^2,y=\cos(x)$$
so finally
$$ y = \frac{1}{\sqrt 3 } \to g(y) = \frac{4}{{3\sqrt 3 }}=f(x)$$
$$ y =  - \frac{1}{\sqrt 3 } \to g(y) =  - \frac{4}{{3\sqrt 3 }}=f(x)$$
which clearly shows the maximum.
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