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Here I have this trigonometric equation $\cos 7\theta=\cos 3\theta+\sin 5\theta$.

I approached the problem as follows : $\cos 7\theta=\cos 3\theta+\sin 5\theta$ $\implies \cos 7\theta-\cos 3\theta=\sin 5\theta$ $\implies 2\sin 5\theta\sin(-2\theta)=\sin 5\theta$

From there on we get, $\sin 2\theta=-\dfrac{1}{2}$ and $\sin 5\theta=0$

So, we deduce that $\theta=-\dfrac{\pi}{12}+\pi k$ and $\theta=\dfrac{2}{5}\pi n$ where $k$ and $n$ are integers.

However, Wolfram|Alpha doesn't seem to agree with me. They present $\dfrac{\pi}{5}$ as a solution which is not included in my solution.

Can someone help me out?

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    $\begingroup$ Note that $\sin x=0$ has solutions $x=n\pi$ (not $x=2n\pi$). $\endgroup$
    – SMM
    Sep 14 '18 at 14:00
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    $\begingroup$ Also note that $\sin x=-\frac{1}{2}$ has solutions $x=-\frac{\pi}{6}+2n\pi$, but also $x=-\frac{5\pi}{6}+2n\pi$. $\endgroup$
    – SMM
    Sep 14 '18 at 14:02
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$$\sin 5\theta=0$$ $\sin x =0$ has solutions $x=n\pi$

Thus we get $$5\theta=n \pi$$

$$\theta= \frac{n\pi}{5}\hspace{10pt} n\in Z$$

Also $$\sin 2\theta=-\dfrac{1}{2}$$

$$2\theta=2n\pi-\frac{\pi}{6} \implies \theta=n\pi-\frac{\pi}{12}\hspace{10pt} n\in Z$$ $$2\theta=(2n+1)\pi+\frac{\pi}{6} \implies \theta=n\pi+\frac{7\pi}{12}\hspace{10pt} n\in Z$$

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