1+sinx+sin^2x+.....=2√3+4, then x=?

We have:
#sum_(n=0)^oosin^n(theta)=2sqrt(3)+4#

#sum_(n=0)^oo(sin(theta))^n=2sqrt(3)+4#

We can try each of these values, and see which gives #2sqrt3+4#

#f(r)=sum_(n=0)^oor^n=1/(1-r)#

#f((3pi)/4)-=f(pi/4)=1/(1-sin(pi/4))=2+sqrt2#

#f(pi/6)=1/(1-sin(pi/6))=2#

#f(pi/3)=1/(1-sin(pi/3))=2sqrt3+4#

#pi/3-=3#

There's another way, using Geometric progression.

The series is #1+sintheta+(sintheta)^2+(sintheta)^3+....+oo# which can be written as
# (sintheta)^0+sintheta+(sintheta)^2+(sintheta)^3+....+oo# #[because "anything"^0=1]#

Our First term of progression #a=1# and common ratio between each term of the series is #r=sintheta#

Sum of an infinite Geometric Progression series is given by :

#S_oo=a/(1-r), r≠1#

Plugging in the values we've

#S_oo=1/(1-sintheta)#

But, #S_oo=2sqrt3+4# is given.
So,

#1/(1-sintheta)=2sqrt3+4#

#=>1/(2sqrt3+4)=1-sintheta#

Rationalising the denominator on Left hand side,

#=>color(red)((2sqrt3-4))/((2sqrt3+4)color(red)((2sqrt3-4))) = 1-sintheta#

#=> (2sqrt3-4)/(12-16)=1-sintheta# #[because (a+b)(a-b)=a^2+b^2]#

#=>-(2sqrt3-4)/4=1-sintheta#

#=>-(cancel2sqrt3)/cancel4^2+4/4=1-sintheta#

#=> -sqrt3/2+cancel1=cancel1-sintheta#

#=>cancel-sqrt3/2=cancel-sintheta#

#=> sqrt3/2=sintheta#

#=>theta=sin^(-1)( sqrt3/2)#

#=> theta=60°=π/3#

Hope this helps. :)