At first I thought it was trivial to lớn bring this in an IMO, but I realized approaching it directly brings a power of 6 which is not all too friendly.

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Is there a sneaky way to lớn solve sầu this?


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Use double angles to lớn get the equivalent equation $$cos(0x) + cos(2x) + cos(4x) + cos(6x) = 0.$$

Pair (first & last term) và (middle ones) to lớn get the equivalent form$$cos(x)cos(2x)cos(3x) = 0.$$


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Use double angle identify to write the equation as

$$cos 2x+cos 4x +cos 6x+1=0$$

Let $t=cos 2x$. Then, the equation is

$$2t^3+t^2-t=0implies t(t+1)(2t-1)=0$$

Thus, $cos 2x= 0,> -1, > frac12$ & the solutions are $x = fracpi4+fracnpi2,> fracpi2+ npi,> pmfracpi6+frac2npi3$.


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Let $t:=cos^2x$.

We have

$$t+(2t-1)^2+(4t-3)^2t=1$$

or

$$16t^3-20t^2+6t=0.$$

The roots are $0,dfrac12,dfrac34$, nothing really difficult.

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Use Prove that $cos (A + B)cos (A - B) = cos ^2A - sin ^2B$

$$cos^2x+cos^22x=cos^22x-sin^2x+1=cos(2x+x)cos(2x-x)+1$$

Finally by http://bigbiglands.comworld.wolfram.com/ProsthaphaeresisFormulas.html

$$cos3x+cos x=2cos2xcos x$$


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It"s $$cos^2xleft(1+(4cos^2x-3)^2 ight)=4sin^2xcos^2x,$$ which gives $$cosx=0$$ and $$x=90^circ+180^circk,$$ where $k$ is an integer number, or $$1+(4cos^2x-3)^2=4-4cos^2x,$$ which is$$8cos^4x-10cos^2x+3=0$$ or$$(4cos^2x-3)(2cos^2x-1)=0$$ or$$(2cos2x-1)cos2x=0,$$ which gives$$x=pm30^circ+180^circk$$ or $$x=45^circ+90^circk.$$


Hint:use the identity $$displaystyle cos ^2alpha +cos ^2eta +cos ^2gamma =-2cos altrộn cos eta cos gamma +1,$$


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