4
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I simplify it to $\cos x(\cos (2x)+\cos (3x))=-\frac{1}{4}$ but don't know how to go further.

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    See http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – 2017-01-18

2 Answers 2

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The identity $\cos a \sin b = \dfrac 12 \bigg( \sin(a+b) - \sin(a-b) \bigg)$ is applicable here. Multiply both sides by $\sin \frac x2$ to get $$\sum_{k=1}^4 \cos kx \sin \frac x2 = - \frac 12 \sin \frac x2$$ and apply the identity: $$\frac 12 \sum_{k=1}^4 \left( \sin (k+ \frac 12)x - \sin(k - \frac 12)x \right) = - \frac 12 \sin \frac x2.$$ The sum on the left telescopes and you end up with $$\sin \frac 92 x - \sin \frac x2 = - \sin \frac x2.$$ This reduces to $$\sin \frac 92 x = 0$$ so that $$x = \frac{2n\pi}{9}.$$

EDIT: a solution to $$\sum_{k=1}^4 \cos kx \sin \frac x2 = - \frac 12 \sin \frac x2$$ is a solution to either $$\sum_{k=1}^4 \cos kx= - \frac 12 \quad \text{or} \quad \sin \frac x2 = 0.$$

A number $x$ is solution to the second equation if and only if $x = 2k\pi$, $k \in \mathbb N$, and moreover no number of this form is a solution to the first equation since the sum would necessarily equal $4$. Thus the solutions to the original equation are $$x = \frac{2n\pi}{9},\quad n \in \mathbb N,\quad n \not\equiv 0 \bmod 9.$$

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    The number of solutions in $[0,2\pi]$ is simple to count at this point. – 2017-01-18
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    Doesn't 0 come out as a solution this way? – 2017-01-18
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    No. That was introduced by the multiplying by $\sin(x/2)$. – 2017-01-18
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    What if $9$ divides $n$ – 2017-01-18
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    Alright, cool, but how does one even get this idea? It doesn't seem like a common trick to me. Thanks anyways! – 2017-01-18
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    Edited. Thanks to John Doe and lab bhattacharjee for pointing out an error. – 2017-01-18
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    @JohnDoe multiplying by $\sin \frac x2$ and telescoping is a well-known trick for obtaining the closed form of a sum of cosines. – 2017-01-18
2

You can also keep going with your idea:

$$\cos x (\cos 2x + \cos 5x)=-1/4 \rightarrow \cos(x/2)\cos(x)\cos(5x/2)=-1/8$$

Now multiply both sides by $\sin (x/2)$:

$$\sin(x/2)\cos(x/2)\cos(x)\cos(5x/2)=(-1/8)\sin(x/2)$$

$$\sin(x)\cos(x)\cos(5x/2)=(-1/4)\sin(x/2)$$

$$\sin(2x)\cos(5x/2)=(-1/2)\sin(x/2)$$

$$\sin(9x/2)-\sin(x/2)=-\sin(x/2)\rightarrow \sin(9x/2)=0$$

$$9x/2=k\pi \rightarrow x= \frac{2k\pi}{9}$$

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    Insightful, thank you :) – 2017-01-19
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    You are welcome – 2017-01-19