<s>[INST] Finding all solutions to a trigonometric equation I was solving a simple trigonometric equation for my brother, going this path:
$16\sin^2(2x)\cos^2(2x) = 3$
${[4\sin(2x)\cos(2x)]}^2 = 3$
Applied the formula for $\sin(2a) = 2\sin(a)\cos(a)$ "backwards"
$[2\sin(4x)]^2 = 3$ 
$\sin(4x) = \frac{\sqrt{3}}{2}$
Then the solutions are easily found:
$\frac{\pi}{12} + k\frac{\pi}{2}$ and $\frac{\pi}{6} + k\frac{\pi}{2}$
However the book lists 4 different solutions. I can find them with little effort, but very arbitrarily. How can I know in advance how many solutions there are to such an equation? And to a generic $f(x) = 0$?
 [/INST] Due to taking the square root, $$\sin(4x)=\pm\frac{\sqrt3}2\implies 4x=\frac\pi3, \frac{2\pi}3, \frac{4\pi}3, \frac{5\pi}3, \frac{7\pi}3, \frac{8\pi}3, \frac{10\pi}3, \frac{11\pi}3$$ so $$\boxed{x=\frac\pi{12}, \frac\pi6, \frac\pi3, \frac{5\pi}{12} \frac{7\pi}{12},\frac{2\pi}3, \frac{5\pi}6, \frac{11\pi}{12}}$$ are the eight solutions for $0<x<2\pi$.
As commented below an infinite set of solutions can be found by adding $\dfrac{k\pi}4$ to each solution for $k\in\mathbb{Z}$.
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