<s>[INST] What is the maximum value of $ \sin x +\sin {2x} (0I would appreciate if somebody could help me with the following problem
Q: What is the maximum value of $$ \sin x +\sin {2x} (0<x<\frac{\pi}{2})$$
I have done my work here 
$$f (x)= \sin x +\sin {2x} $$
$$f'(x)= \cos x +2\cos {2x}=\cos x + 4\cos^2 x-2 =0$$
$$\cos x=\frac{1}{8} \left(\sqrt{33}-1\right),\cos x=\frac{1}{8} \left(-\sqrt{33}-1\right)$$
Done...!
 [/INST] $$f (x)= \sin x +\sin {2x} $$
$$f'(x)= \cos x +2\cos {2x}=\cos x + 4\cos^2 x-2 = 0 \implies
\cos x = \frac{-1 \pm \sqrt {33}}{8}$$
Since we are interested in the maxima only, discard the negative root. In general, $$x = 2k\pi \pm \cos^{-1}\frac{\sqrt{33}-1}{8} = 2k\pi \pm 0.93592946$$
However, in this case $$x = 0.93592946$$
Since $$f''(x) = -(\sin x + 4 \sin 2x)$$ verify that for the above value of $x$, you really have $f''(x) < 0$
In fact, $$f''(0.93592946) = -4.62524$$ and the maxima is $$f(0.93592946) = 1.76017$$
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