First slide

Trigonometric equations

Question

If $1 + \sin x + \sin^{2} x + \sin^{3} x + \dots + \dots \infty$ is equal to $4 + 2 \sqrt{3},$ $0 < x < π,$ then $x =$

Moderate

A

$\frac{π}{6}$
B

$\frac{π}{4}$
C

$\frac{π}{3} or \frac{π}{6}$
D

$\frac{π}{3} or \frac{2 π}{3}$

Solution

We have,
$0 < x < π \Rightarrow \sin x > 0$
Now,
$\begin{array}{l} 1 + \sin x + \sin^{2} x + \dots x = 4 + 2 \sqrt{3} \\ \Rightarrow \frac{1}{1 - \sin x} = 4 + 2 \sqrt{3} \\ \Rightarrow \sin x = 1 - \frac{1}{4 + 2 \sqrt{3}} \\ \Rightarrow \sin x = \frac{3 + 2 \sqrt{3}}{4 + 2 \sqrt{3}} = \frac{\sqrt{3}}{2} \Rightarrow x = \frac{π}{3} or, \frac{2 π}{3} \end{array}$

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