Question

If 1 + sin \(x\) + sin² \(x\) + sin³ \(x\) + ..... ∞ = 4 + 2√3 , 0 < \(x\) < π, then find the value of \(x\).

Answer 1

Given, 1 + sin\(x\) + sin²\(x\) + sin³\(x\) + ..... ∞ = 4 + 2√3

⇒ \(\frac{1}{1-\text{sin }x}\) = 4 + 2√3

⇒ 1 - sin \(x\) = \(\frac{1}{4+2\sqrt3}\)

⇒ sin \(x\) = 1 - \(\frac{1}{4+2\sqrt3}\) = \(\frac{4+2\sqrt3-1}{4+2\sqrt3}\) = \(\frac{3+2\sqrt3}{4+2\sqrt3}\) x \(\frac{4-2\sqrt3}{4-2\sqrt3}\)

= \(\frac{12+8\sqrt3-6\sqrt3-12}{16-12}\) = \(\frac{2\sqrt3}{4}\) = \(\frac{\sqrt3}{2}\)

⇒ \(x\) = \(\frac{\pi}{3}\) or \(\frac{2\pi}{3}\) as 0 < \(x\) < \(\pi\).