<s>[INST] Maximum of a trigonometric function without derivatives I know that I can find the maximum of this function by using derivatives but is there an other way of finding the maximum that does not involve derivatives? Maybe use a well-known inequality or identity?
$f(x)=\sin(2x)+2\sin(x)$ 
 [/INST] The idea is to use $\sin^2 x + \cos^2x = 1$ to reduce to dealing with only 1 trigonometric function, and then proceed as a standard 1-variable inequality.
We wish to find the maximum of $f(x) = \sin 2x + 2 \sin x = 2 \sin x ( 1 + \cos x )$.
It is clear that we may assume $\sin x \geq 0, \cos x \geq 0$, to maximize this product.
Let's consider $$[f(x)]^2 = 4 \sin^2 x ( 1 + \cos x)^2 = 4 (1-\cos^2x ) ( 1 + \cos x)^2 = 4 ( 1 - \cos x ) ( 1 + \cos x)^3 $$
By AM-GM, applied to $ 3(1-\cos x) , (1 + \cos x), (1 + \cos x), (1 + \cos x)$, we get that
$ \sqrt[4]{3 ( 1 - \cos x ) ( 1 + \cos x)^3} \leq \frac{ 6}{4}$, or that $ ( 1 - \cos x ) ( 1 + \cos x)^3 \leq \frac{27}{16}$.
Hence, $ [f(x)]^2 \leq \frac{27}{4}$ so $f(x) \leq \frac{3 \sqrt{3} } {2} $.
It remains to verify that equality can occur, which it does at $3(1-\cos x) = 1 + \cos x$ of $\cos x = \frac{1}{2}$.
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