You can put this solution on YOUR website! 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
