If `1+sinx+sin^2x+sin^3x+oo` is equal to `4+2sqrt(3),0<x<pi,` then `x` is equal to `pi/6` (b) `pi/4` (c) `pi/3orpi/6` (d) `pi/3or(2pi)/3`
If `1+sinx+sin^2x+sin^3x+oo`
is equal to `4+2sqrt(3),0A. `pi/6`
B. `pi/4`
C. `pi/3or pi/6`
D. `pi/3 or (2pi)/3`
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Correct Answer - D
Since `0ltxltpi." Therefore "sinxgt0`
We have `1+sinx+sin^2x+…oo=4+2sqrt3`
`rArr 1/(1-sinx)=4+2sqrt3` (sum of infinite G.P.)
`or sinx=1-1/(4+2sqrt3)=(3+2sqrt3)/(4+2sqrt3)=sqrt3/2`
`rArr x=pi/3or(2pi)/3`
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