Question

Find the value of \( x \) satisfying: \[ 1 + \sin x + \sin^2 x + \sin^3 x + \dots = 4 + 2\sqrt{3} \] where \( 0x\pi, x \neq \frac{\pi}{2} \).

• A. \( \frac{\pi}{6}, \frac{\pi}{4} \)
• B. \( \frac{\pi}{4}, \frac{5\pi}{6} \)
• C. \( \frac{2\pi}{5}, \frac{\pi}{6} \)
• D. \( \frac{\pi}{3}, \frac{2\pi}{3} \)

Hint:

For infinite geometric series, use \( S = \frac{a}{1 - r} \) and solve algebraically.

Answer 1

The Correct Option is D

Step 1: Recognizing an Infinite Geometric Series
The given equation: \[ 1 + \sin x + \sin^2 x + \sin^3 x + \dots = S \] is an infinite geometric series with first term \( a = 1 \) and common ratio \( r = \sin x \): \[ S = \frac{1}{1 - \sin x} \] Step 2: Solving for \( x \)
\[ \frac{1}{1 - \sin x} = 4 + 2\sqrt{3} \] \[ 1 - \sin x = \frac{1}{4 + 2\sqrt{3}} \] Rationalizing the denominator: \[ 1 - \sin x = \frac{4 - 2\sqrt{3}}{10} = \frac{2 - \sqrt{3}}{5} \] \[ \sin x = 1 - \frac{2 - \sqrt{3}}{5} = \frac{3 + \sqrt{3}}{5} \] Comparing values, we find: \[ x = \frac{\pi}{3}, \frac{2\pi}{3} \] Thus, the correct answer is \( \frac{\pi}{3}, \frac{2\pi}{3} \).