The general solution of the equation 1−sinx+....+−1nsinnx+....1+sinx+....+sinnx+.....=1−cos2x1+cos2x , x≠2n+1π/2,n∈I is
−1nπ/3+nπ
−1nπ/6+nπ
−1n+1π/6+nπ
−1n−1π/3+nπ
We have 1−sinx+....+−1nsinnx+....1+sinx+....+sinnx+.....=1−cos2x1+cos2x
Numerator and denominator both are in GP S∞=a1-r
⇒11+sinx×1−sinx1=2sin2x2cos2x,∵−1<sinx<1
⇒1−sinx=sin2x1+sinx1−sin2x
⇒1−sinx2=sin2x⇒1−2sinx=0
⇒sinx=12=sinπ6
⇒x=nπ+−1nπ6