write cosx + cos3x = 2cos(2x)cos(x)





it becomes....



cos(2x) + 2cos(2x)cos(x)=cos(2x)[1+2cos(x)]=1/4



cos(2x)[1+2cos(x)]=cos(2x)sin^2(x)=1/8



(1-2sin^2x)sin^2x=1/8



sin^2x - 2sin^4x = 1/8



let sin^2x=t



t-2t^2=1/8



8t-16t^2=1.....solve this eqn for t..u get....t=1/4...now frame the general solution :)