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4cosx cos2x cos3x=1,if 0
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What's that, solve or prove true?
Cos nx = 2 cos x cos(n-1)x -cos(n-2)x
2(2cosxcos3x)cos2x=1 2(cos2x+cos4x)cos2x=1 2(cos2x)^2+2cos2xcos4x=1 2(cos2x)^2+cos6x+cos2x=1\[2\cos ^{2}2x-1+\cos6x+\cos2x=0\] cos4x+cos6x+cos2x=0 cos4x+2cos4xcosx=0 cos4x(1+2cosx)=0 cos4x=0 or cosx=-1/2 x=90 or x=120
4cosx cos 2x cos3x=1 (2cosx cos3x)(2cos2x)=1 (cos4x+cos2x)(2cos2x)=1 2cos4x cos2x+2cos^2 2x-1=0 2cos4x cos2x+cos4x=0 cos4x(2cos2x+1)=0 cos4x=0 2cos2x+1=0
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Thats what exactly i have written.............thanks for verifying my answer
thanks to u man,i was just checking it out
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