If 1+ sin x+ sin 2 x+ upto ∞ =4+2 √3, 0

Question

If 1+ sin x+ sin 2 x+ upto ∞ =4+2 √3, 0<x<π and x ≠ (π/2), then x is equal to (A)  (π /3),(5π /6)  (B)  (2π /3),(π /6)  (C)  (π /3),(2π /3)  (D)  (π /6),(π /3)

Tardigrade Admin · Accepted Answer

Given, 1+ sin x+ sin 2 x+... ∞=4+2 √3 ⇒ (1/1- sin x)=4+2 √3 ⇒ 1- sin x=(1/4+2 √3) × (4-2 √3/4-2 √3) (4-2 √3/4) ⇒ sin x =(2 √3/4)=(√3/2) ⇒ x =(π/3) (2 π/3)

Q. If $1 + sin x + sin^{2} x +$ upto $\infty$ $= 4 + 23, 0 < x < π$ and $x \neq = \frac{π}{2}$ , then $x$ is equal to

$\frac{π}{3}, \frac{5 π}{6}$

$\frac{2 π}{3}, \frac{π}{6}$

$\frac{π}{3}, \frac{2 π}{3}$

$\frac{π}{6}, \frac{π}{3}$

Given, $1 + sin x + sin^{2} x + ...\infty = 4 + 23$
$\Rightarrow \frac{1}{1 - s i n x} = 4 + 23$
$\Rightarrow 1 - sin x = \frac{1}{4 + 2 3} \times \frac{4 - 2 3}{4 - 2 3}$
$\frac{4 - 2 3}{4}$
$\Rightarrow sin x = \frac{2 3}{4} = \frac{3}{2}$
$\Rightarrow x = \frac{π}{3} \frac{2 π}{3}$

Q. If 1+sinx+sin2x+ upto ∞ =4+23​,0<x<π and x=2π​, then x is equal to

3π​,65π​

32π​,6π​

3π​,32π​

6π​,3π​