<s>[INST] Find all extrema of a complicated trigonometric function Problem
Find all local extrema for
$$f(x) = \frac{\sin{3x}}{1+\frac{1}{2}\cos{3x}}$$
Attempted solution
My basic approach is to take the derivative, set the derivative equal to zero and solve for x.
Taking the derivative with the quotient rule and a few cases of the chain rule for the trigonometric functions with a final application of the Pythagorean identity:
$$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})+1.5\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2} = \frac{3\cos 3x+1.5\cos^2 3x + 1.5\sin^2 3x}{(1+\frac{1}{2}\cos 3x)^2} = \frac{3 \cos 3x + 1}{(1+\frac{1}{2}\cos 3x)^2}$$
Putting it equal to zero and solving for x:
$$3\cos 3x + 1 = 0 \Rightarrow x = \frac{\arccos{\Big(-\frac{1}{3}\Big)}}{3} = \frac{\pi}{6} + \frac{2\pi n}{3}$$
...however the expected answer is $\pm\frac{2\pi}{9} + \frac{2\pi n}{3}$
So I must have gone wrong somewhere.
 [/INST] It should be $$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})-\frac{3}{2}\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2}.$$ 
I like the following way.
Let $x=\frac{2\pi}{9}.$
Thus, we get a value $\frac{2}{\sqrt3}.$
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\frac{\sin3x}{2+\cos3x}\leq\frac{1}{\sqrt3}$$ or
$$\sqrt3\sin3x-\cos3x\leq2,$$ which is true by C-S:
$$\sqrt3\sin3x-\cos3x\leq\sqrt{((\sqrt3)^2+(-1)^2)(\sin^23x+\cos^23x)}=2.$$
By the same way we can get a minimal value.
I got $-\frac{2}{\sqrt3},$ which  occurs for $x=-\frac{2\pi}{9}.$
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