# Trigonometric Equation Problem

I need to solve the following equation: $(-12\cos^2(x)-2\cos(x)+3)(-\sin(x))=0$ on the interval $(0,2\pi)$

I found 3 out of 5 solutions which are the following: $\pi, \arccos\left(\frac{-1-\sqrt{37}}{12}\right) \mbox{ and } \arccos\left(\frac{\sqrt{37}-1}{12}\right)$

Can someone help me find the 2 other solutions

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Why do you believe there are 5 solutions? You have a trigonometric polynomial of degree 3 limited to $\,(0,2\pi)\,$, so three solutions look good for me...unless you've some further info. – DonAntonio Oct 29 '12 at 17:41
I see on the graph that there exists two additional solutions – user43418 Oct 29 '12 at 17:42
Oh, I see...... – DonAntonio Oct 29 '12 at 17:46

Observe that $\cos x=\cos (2\pi-x)$. So if $\alpha$ is a solution, then so is $2\pi-\alpha$.
So you have $\pi, \arccos\left(\frac{-1-\sqrt{37}}{12}\right), \arccos\left(\frac{-1+\sqrt{37}}{12}\right),2\pi-\arccos\left(\frac{-1-\sqrt{37}}{12}\right),2\pi-\arccos\left(\frac{-1+\sqrt{37}}{12}\right)$
5 solutions, as required.

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Thank you Mr. Gulko – user43418 Oct 29 '12 at 18:14
@user43418: You're welcome :-) – Dennis Gulko Oct 29 '12 at 18:15