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Solve the equation $$\cos \left(x+30^{\circ}\right)+\cos \left(x+10^{\circ}\right)=\cos \left(2x+10^{\circ}\right)+\cos 10^{\circ}$$ , where $x\in (0,\pi )$

My try: $$\cos{(x+30)}+\cos{(x+10)}=2\cos{(x+20)}\cos{10}$$

and $$\cos{(2x+10)}+\cos{10}=2\cos{(x+10)}\cos{x}$$ then $$\cos{(x+20)}\cos{10}=\cos{(x+10)}\cos{x}$$

then I can't work,Thank you

math110
  • 93,304
  • This can be solved by one of the iterative methods in Numerical Analysis.

    Using some calculus for finding points of extrema, one still has to fall back on Numerical Analysis.

    Perhaps there is some trick in Trigonometry but it fails me right now!

     – wannadeleteacct Oct 23 '13 at 12:48

1 Answers1

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I think it's prohibitively difficult solve such an equation like this one in a standard way as we can not take out any common factor.

So, I have tried to set one of cosine term $=0$ in one side and check. Eventually set the other side will have to be of the form $\cos y=2\cos60^\circ\cos y=\cos(60^\circ-y)+\cos(60^\circ+y)$

For example,

if we set $\cos(x+30^\circ)=0, x+30^\circ=(2n+1)90^\circ\iff x=180^\circ n+60^\circ,$ where $n$ is any integer

Observe that $x=360^\circ m+60^\circ$ satisfies the given equation, where $m$ is any integer

Check for $\cos(x+10^\circ)=0$ and $\cos(2x+10^\circ)=0$

I have validated from Wolfram Alpha that $40^\circ,60^\circ$ are the only two solutions within $(0,\pi)$

lab bhattacharjee
  • 274,582
  • @math110, another way could be to equate $2x+10^\circ, x+30^\circ$ and $x+10^\circ$ one by one with $60^\circ$ here(http://math.stackexchange.com/questions/537275/relationship-among-a-b-c-d-for-cos-a-cos-b-cos-c-cos-d) – lab bhattacharjee Oct 23 '13 at 19:03