Question

Sin2x+sin4x+sin6x=0 general values of it

Answer 1

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sin2x + sin4x + sin6x =0

(sin2x + sin6x) + sin4x=0

2sin4x.cos2x + sin4x=0

2sin4x.(2cos2x + 1)=0

2sin4x.[2(2cos^2x-1)+1]=0

2sin4x.[4cos^2x-1]=0

hence;

sin4x = 0 or 4cos^2x-1 = 0

sin4x = sin0 or 4cos^2x = 1

4x =0 or 2cos x = +1 or 2cos x = -1

4x = nπ or cos x = 1/2 or cos x = -1/2

x = nπ/4 or cos x = cos π/3 or cos x = -cosπ/3

x = nπ/4 or x= 2mπ+-π/3 or cos x = cos(π-π/3)

x = nπ/4 or x= 2mπ+-π/3 or cos x = cos 2π/3

x = nπ/4 or x= 2mπ+-π/3 or x = 2kπ+-2π/3

n,m,k are Constant.

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