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Trigonometrical Equations
medium
If $1 + \sin x + {\sin ^2}x + .....$ to $\infty = 4 + 2\sqrt 3 ,\,0 < x < \pi ,$ then
A
$x = \frac{\pi }{6}$
B
$x = \frac{\pi }{3}$
C
$x = \frac{\pi }{3}$ or $\frac{\pi }{6}$
D
$x = \frac{\pi }{3}$ or $\frac{{2\pi }}{3}$
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 $ $x = \frac{\pi }{3}$ or $\frac{{2\pi }}{3}$.
Standard 11
Mathematics
