vikas askiitian expert
Last Activity: 13 Years ago
sinxsin3x = 1/2
2sinxsin3x = 1
cos-2x - cos4x = 1 (2sinasinb = cosa-b - cosa+b)
cos2x - cos4x = 1 ...........1
cos4x = 2cos22x - 1 so eq 1 becomes
cos2x - (2cos22x-1) = 1
cos2x - 2cos22x = 0
cos2x(1-2cos2x) = 0
cos2x = 0 or 1-2cos2x = 0
2x = (2n+1)pi/2 2x = 2npi +(-)pi/3
x = (2n+1)pi/4 or x = npi +(-)pi/6
these are the required solutions of this equation
approve if u like my ans