Trigonometrical Equations
medium

यदि $1 + \sin x + {\sin ^2}x + .....$ $\infty $ तक $ = 4 + 2\sqrt 3 ,\,0 < x < \pi ,$ तो

A

$x = \frac{\pi }{6}$

B

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

C

$x = \frac{\pi }{3}$ या $\frac{\pi }{6}$

D

$x = \frac{\pi }{3}$ या $\frac{{2\pi }}{3}$

Solution

(d) $1 + \sin x + {\sin ^2}x + ….\infty = 4 + 2\sqrt 3 $

$ \Rightarrow $ $\frac{1}{{1 – \sin x}} = 4 + 2\sqrt 3 $

$ \Rightarrow $ $\sin x = 1 – \frac{1}{{4 + 2\sqrt 3 }}$

$ \Rightarrow $ $\sin x = 1 – \frac{{(4 – 2\sqrt 3 )}}{4} = \frac{{2\sqrt 3 }}{4} = \frac{{\sqrt 3 }}{2}$

$ \Rightarrow $ $x = \frac{\pi }{3}$ या $\frac{{2\pi }}{3}$.

Standard 11
Mathematics

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