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
Thus, $cos 2x= 0,> -1, > frac12$ & the solutions are $x = fracpi4+fracnpi2,> fracpi2+ npi,> pmfracpi6+frac2npi3$.
The roots are $0,dfrac12,dfrac34$, nothing really difficult.
Use Prove that $cos (A + B)cos (A - B) = cos ^2A - sin ^2B$
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 $$displaystyle cos ^2alpha +cos ^2eta +cos ^2gamma =-2cos altrộn cos eta cos gamma +1,$$
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Explain the concept behind solving $sin(x)cos(x) + cos(x) = 0$, from Paul's Online bigbiglands.com Notes
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