Solve the trig eq

Solve the equation :
cos²x + cos²2x + cos²3x = 1 for all x

2 Answers

cos^2x=\frac{1+cos2x}{2}
so the eq. in equivalent to

(1+cos2x)+(1+cos4x)+(1+cos6x)=2

or, 3+cos2x+cos6x+cos4x=2

or,(2cos²x-1)+2(cos5x)(cosx)=-1

or,2cos²x+2(cos5x)(cosx)=0

or,cosx(cosx+cos5x)=0

or,cosx{(2)(cos3x)(cos2x)}=0

so, either cosx=0 or cos2x=0 or cos3x=0

Use \cos ^2(x)+ \cos^2 (2x)+ \cos ^2 (3x)-1 = 2\cos^2(x)\cos(2x)(2\cos(2x)-1)

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