# Problem with trigonometric equation

$12(\cos(x))^3+2\cos(x)^2+(24\sin(x)-3)\cos(x)+2\sin(x)= 0$

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What does this have to do with calculus? Also, what have you tried already? – apnorton Oct 29 '12 at 18:25
Well I don't really know where to start.. – user43418 Oct 29 '12 at 18:25
Have you tried factoring by grouping? – Ben Oct 29 '12 at 18:26
What does that mean ? – user43418 Oct 29 '12 at 18:27
So apparently, it isn't possible to determine exact values of x... – user43418 Oct 29 '12 at 18:56

$s = \sin(x)$ must satisfy the equation $144\,{s}^{6}-576\,{s}^{5}+220\,{s}^{4}+1000\,{s}^{3}-283\,{s}^{2}-424 \,s-77 = 0$. This has Galois group $S_6$, so it can't be solved in terms of radicals. Thus you aren't going to get nice closed-form solutions. The four solutions for $0 \le x \le 2 \pi$ are approximately $1.66661701719437, 3.42548142597468, 4.63849563287631, 5.91793801389173$ (found by numerical methods).