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?

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.\$\$

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\$.

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.

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\$\$

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 ^2eta +cos ^2gamma =-2cos altrộn cos eta cos gamma +1,\$\$

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