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Trigonometry Solve cos5x+cosx=cos6x

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MHB POTW Director
Staff member
Feb 14, 2012
Solve $cos5x+cosx=cos6x$ in the interval $0<x<2\pi$.

Without thinking much, I do the following:
Now, even if I still refuse to devise a plan but blindly applying the trig formula, I get:

Hmm, clearly this isn't going to bring me to any credible output.:(

But questions like this often yield the cases where:
1. zero=zero
2. 1 = 1

Thus, without actually solving it, I can say that
$x=\frac{\pi}{4}, \frac{\pi}{3}, and \frac{3\pi}{4}$ in the interval $0<x<2\pi$. I must have left out some other values. That is for sure.

That is my thought.

Despite my having said this, I tried to expand it and get down to all terms involving only $cos^nx$ and/or $sin^nx$

Ah! This is a move in the wrong direction.

Could someone help me, please?



MHB Oldtimer
Staff member
Feb 7, 2012
If you use the multiple angle formulae for $\cos(5x)$ and $\cos(6x)$ and write $c=\cos x$ then the equation becomes $32c^6-16c^5-48c^4+20c^3+18c^2-6c-1=0.$ That factorises as $(2c-1)(2c^2-1)(8c^3-8c-1)=0.$

The first factor $2c-1$ gives the equation $\cos x=1/2$, corresponding to the solutions $x=\pi/3$ and $x=5\pi/3$.

The second factor $2c^2-1$ gives the equation $\cos x=\pm1/\sqrt2$, corresponding to the solutions $x=\pi/4$, $x=3\pi/4$, $x=5\pi/4$ and $x=7\pi/4$.

But the third factor $8c^3-8c-1$ look completely intractable to me. The cubic equation $8c^3-8c-1=0$ has three real roots. One root is greater than 1 and so does not correspond to any values of $\cos x.$ The other two roots are negative, and as far as I can see do not correspond to any recognisable values of $x.$
Last edited:


Well-known member
Feb 13, 2012
Using Euler's identity $\displaystyle \cos x = \frac{e^{i x}+ e^{- i x}}{2}$ the trigonometric equation becomes...

$\displaystyle e^{5 i x} + e^{-5 i x}+ e^{i x} + e^{- i x}= e^{i 6 x} + e^{ -6 i x}$ (1)

... and setting $\displaystyle e^{i x}= \xi$ is some steps we arrive to write...

$\displaystyle \xi^{12} -\xi^{11}- \xi^{7} - \xi^{5} - \xi +1= (\xi^{2}-\xi+1)\ (\xi^{4}+1)\ (\xi^{6}-\xi^{4} -\xi^{3}-\xi^{2}+1)=0$ (2)

Because is $\displaystyle x=-i\ \ln \xi$ the solution of (1) will be derived from the solution of (2) which lie on the unit circle. The factor $\displaystyle \xi^{2} - \xi +1$ conducts us to the solutions $\displaystyle x=\pm \frac{\pi}{3}$. The factor $\displaystyle \xi^{4}+1$ conducts us to the solutions $\displaystyle x= (2n+1)\ \frac{\pi}{4}$.The last factor $\displaystyle \xi^{6} -\xi^{4} -\xi^{3}-\xi^{2}+1$ is of course less 'comfortable' but its numerical solution shows that four of its complex roots lie on the unit circle...


... and are in the second and third quadrants...

Kind regards

$\chi$ $\sigma$