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?
$f(x)=\sin(2x)+2\sin(x)$ 
 A: 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 $\cos x = \frac{1}{2}$.
A: As Calvin Lin has observed $\sin x,\cos x>0 \implies 0\le x\le \frac\pi2$
$\implies 0\le 2x\le \pi\implies0\le \pi-2x\le \pi$
As sine function is concave in $[0,\pi]$
$$\sin2x+2\sin x$$
$$=\sin(\pi-2x)+\sin x+\sin x\le 3\sin\frac{(\pi-2x+x+x)}3=3\sin\frac{\pi}3=\frac{3\sqrt3}2$$  the equality occurs when $x=x=\pi-2x$ i.e., when $x=\frac\pi3$ 
