If 1 + sin x + sin^2 x + dots infty = 4 + 2sq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin x$.For the sum to exist,$|\sin x| < 1$.Th

Vedclass · Accepted Answer

$x = \frac{\pi}{3}$

Vedclass · Answer

(B) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = \sin x$.For the sum to exist,$|\sin x| The sum of the infinite series is given by $S = \frac{a}{1 - r} = \frac{1}{1 - \sin x}$.Given $\frac{1}{1 - \sin x} = 4 + 2\sqrt{3}$.$1 - \sin x = \frac{1}{4 + 2\sqrt{3}} = \frac{4 - 2\sqrt{3}}{(4 + 2\sqrt{3})(4 - 2\sqrt{3})} = \frac{4 - 2\sqrt{3}}{16 - 12} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2}$.Therefore,$\sin x = \frac{\sqrt{3}}{2}$.Since $0 Comparing with the given options,$x = \frac{\pi}{3}$ is a valid solution.

If $1 + \sin x + \sin^2 x + \dots \infty = 4 + 2\sqrt{3}$ for $0 < x < \pi$,then:

Similar Questions

If $x, 2x + 2, 3x + 3$ are in $G.P.$,then the fourth term is

The roots of the equation $x^3-14x^2+56x-64=0$ are in

The first three terms of a geometric progression are $a, b, c$. If the harmonic mean of $a$ and $b$ is $12$ and the harmonic mean of $b$ and $c$ is $36$,then $a = \dots$

The $6^{th}$ term of a $G.P.$ is $32$ and its $8^{th}$ term is $128$,then the common ratio of the $G.P.$ is

The geometric mean of $n$ positive terms $x_1, x_2, \dots, x_n$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module