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I've been practicing trigonometric equations lately and I've stumbled upon this problem

$$\cos x + \cos3x = 2$$

I tried to write $\cos3x$ as $\cos(2x+x)$ and then I got $\cos2x \cdot \cos x - \sin2x\sin x$

Then when I finally make the equations as simple like this:

$$\cos^2x(\cos x - 1) = 2$$

I don't know what to do anymore. Any help would really appreciate it.

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    $\begingroup$ When I put $x=0$ in your given equation, it works. But when I put $x=0$ in your last equation, it does not work. Something ain't right. You can also make a graph of the given equation to see what is going on, it becomes straight forward... $\endgroup$ – imranfat Jul 15 '16 at 18:34
  • $\begingroup$ What I did is took Bernard answer and I tried this. $2cos2x*cosx = 2$ We divide everything by 2 and we get $cos2x*cosx = 1$ Now I made two cases $cos2x = 1$ and $cosx = 1$ That is the first case and the second is $cos2x = -1$ and $cosx = -1$. Is this right ? $\endgroup$ – Gigaxel Jul 15 '16 at 18:53
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This equation implies $\;\cos x=1\;$ and $\;\cos 3x=1$.

Note: That said, the standard way to factor the l.h.s. would be to write $$\cos x+\cos3x=\cos(2x-x)+\cos(2x+x)=2\cos x\cos 2x,$$ but it's useful only if the r.h.s. is $0$.

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  • $\begingroup$ What I did is took your answer and I tried this. $2cos2x*cosx = 2$ We divide everything by 2 and we get $cos2x*cosx = 1$ Now I made two cases $cos2x = 1$ and $cosx = 1$ That is the first case and the second is $cos2x = -1$ and $cosx = -1$. Is this right ? $\endgroup$ – Gigaxel Jul 15 '16 at 18:53
  • $\begingroup$ Except the second case can't happen: if $\cos x=-1$, $\,\cos 2x=+1$. $\endgroup$ – Bernard Jul 15 '16 at 18:59
  • $\begingroup$ For several reasons: one of them is that $\cos x=-1$ means $x$ is an odd multiple of $\pi$ In such a case, $2x$ is an even multiple of $\pi$. Another reason is that $\;\cos 2x=2\cos^2x-1$. $\endgroup$ – Bernard Jul 15 '16 at 19:10
  • $\begingroup$ So the solution is only $cos2x = 1$ and $cosx = 1$ ? $\endgroup$ – Gigaxel Jul 15 '16 at 19:12
  • $\begingroup$ Absolutely. Anyway, the hint is much simpler, as it uses only $\cos x\le 1$. $\endgroup$ – Bernard Jul 15 '16 at 20:04
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If you go in your way solving, you should obtain $\cos{3x}=4\cos^3{x}-3\cos{x}$.

Therefore, equation becomes $$4\cos^3{x}-2\cos{x}-2=0,$$ $$2(\cos^3{x}-\cos{x})+2(\cos^3{x}-1)=0,$$ $$2(\cos{x}-1)(2\cos^2{x}+2\cos{x}+1)=0.$$

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  • $\begingroup$ $2cos2x*cosx = 2$ We divide everything by 2 and we get $cos2x*cosx = 1$ Now I made two cases $cos2x = 1$ and $cosx = 1$ That is the first case and the second is $cos2x = -1$ and $cosx = -1$. Is this also right ? $\endgroup$ – Gigaxel Jul 15 '16 at 18:59
  • $\begingroup$ @Gigaxel If $\cos{x}=-1$, then $\sin{x}=0$ from identity $\sin^2{x}+\cos^2{x}=1$ and also $\cos{2x}=\cos^2{x}-\sin^2{x}=1$. Therefore, second case isn't possible. $\endgroup$ – alans Jul 15 '16 at 19:06
  • $\begingroup$ So the solution is only $cos2x = 1$ and $cosx = 1$ ? $\endgroup$ – Gigaxel Jul 15 '16 at 19:09
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    $\begingroup$ @Gigaxel Yes, but also if $\cos{x}=1$, we have $\cos{2x}=2\cos^2{x}-1=2-1=1$. So, intersection of these two is $\cos{x}=1$. You can also conclude that from my solution. Final solution is $x=2n\pi$ for $n\in Z$ $\endgroup$ – alans Jul 15 '16 at 19:16

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