Question

1+sinx+sin^2x+.....=2√3+4, then x=?

Answer 1

`3:` `pi/3`

Explanation:

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`

Answer 2

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. :)