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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 $f(x) = 0$?

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  • $\begingroup$ I noticed that the exponent are 2 for sine and 2 for cosine and couldn't help but think about the fundamental theorem of algebra, but this is not a polynomial by any means. Or do I need to consider for example: $t=sin(x)$ and $q=cos(x)$ with which it becomes a 4th degree polynomial? $\endgroup$
    – user258607
    Feb 1, 2018 at 20:19
  • $\begingroup$ $4\sin^2 4x = 3 \implies \sin 4x = \pm \frac {\sqrt {3}}{2}$ which has $4$ solutions in $[0,2\pi]$ $\endgroup$
    – Doug M
    Feb 1, 2018 at 20:37
  • $\begingroup$ Is the title correct? $\endgroup$ Feb 1, 2018 at 20:40
  • $\begingroup$ @JaideepKhare Not far off though ;) $\endgroup$
    – TheSimpliFire
    Feb 1, 2018 at 20:44

4 Answers 4

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You have missed that $[2\sin(4x)]^2=3$ can also mean $\sin(4x)=-\frac{\sqrt3}2$. And being able to see how many solutions there are comes mostly from experience, and it is not infallible.

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use that $$2\sin(2x)\cos(2x)=\sin(4x)$$ and $$\sqrt{(2\sin(4x))^2}=|2\sin(4x)|$$

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  • $\begingroup$ OP already has that. $\endgroup$
    – TheSimpliFire
    Feb 1, 2018 at 20:21
  • $\begingroup$ ok and he has forgotten the absolute value signs $\endgroup$ Feb 1, 2018 at 20:23
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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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  • $\begingroup$ it must be $$\sin(4x)=\pm\frac{\sqrt{3}}{2}$$ and nothing else $\endgroup$ Feb 1, 2018 at 20:24
  • $\begingroup$ and so you will get not all Solutions! $\endgroup$ Feb 1, 2018 at 20:25
  • $\begingroup$ ok errare humanum est! $\endgroup$ Feb 1, 2018 at 20:26
  • $\begingroup$ There's an infinite number of solutions, like for all trigonometric equations. $\endgroup$
    – Bernard
    Feb 1, 2018 at 21:06
  • $\begingroup$ @Bernard Added this. $\endgroup$
    – TheSimpliFire
    Feb 2, 2018 at 8:09
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A shorter presentation of the results: $$\sin 4x=\pm\frac{\sqrt 3}2\iff 4x\equiv \frac\pi3,\;\frac{2\pi}3 \mod \pi\iff x\equiv \frac{\pi}{12}\;\frac{\pi}{6}\mod \frac\pi 4. $$

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