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Solve $x$ in the equation $2\cos3x + \cos2x + 1= 0$, where $0 \leq x \leq 2\pi$.

I tried breaking up the triple angle, into a cos compound angle. But, im not sure what to do next.

$2\cos x \cos2x - 2\sin x \sin2x + \cos2x+1=0$

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    $\begingroup$ If you keep going you can obtain a cubic in $\cos(x)$. Specifically, use $\sin(2x)=2\sin(x)\cos(x)$ and $\cos(2x) = 2\cos^2(x)-1$ $\endgroup$
    – David P
    Apr 26 '14 at 23:21
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Using Prove $\cos 3x =4\cos^3x-3\cos x$ and Double-Angle Formula, we get $$2(4c^3-3c)+2c^2-1+1=0\iff2c(4c^2+c-3)=0$$ where $c=\cos x$

Can you take it from here?

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  • $\begingroup$ @user146001, How about this? $\endgroup$ May 1 '14 at 12:14

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