If 1+sinx+sin2x+sin3x+…∞ is equal to 4+23,0<x<π then x is equal to
π6
π4
π3 or π6
π3 or 2π3
Since 0 < x < π. Therefore, sinx > 0
We have 1+sinx+sin2x+…∞=4+23
⇒ 11−sinx=4+23 (sum of infinite G.P.)
or sinx=1−14+23=3+234+23=32⇒ x=π3 or 2π3