Question 1062749
as you know it is infinite series so
1+sinx+sin^2x+sin^3x + ..... = 4 + 2√3     eq(1)
 let's common "sinx" from left side 
 
1 + sinx (1+sinx+sin^2x+sin^3x + ..... ) = 4 + 2√3

1 + sinx ( 4 + 2√3 ) = 4 + 2√3       by eq(1)

 by simplifying we get 
 
  sinx = 1 - 1 /(4 + 2√3)
  sinx = (3+2√3)/(4 + 2√3)       
  sinx = √3 /2                             ( put in calculator )

  by taking sine inverse of right side we get

    x = 60