Question:
What is/are the solution(s) of the trigonometric equation cosec x + cot x = \(\sqrt{3}\), where \(0 < x < 2\pi\)?
त्रिकोणमितीय समीकरण \(\operatorname{cosec} x + \cot x = \sqrt{3}\), जहाँ \(0 < x < 2\pi\) है, का/के हल क्या है/हैं?
Options:
- \(\pi, \dfrac{\pi}{3}, \dfrac{5\pi}{3}\)
- \(\dfrac{5\pi}{3}\) only
- \(\pi\) only
- \(\dfrac{\pi}{3}, \dfrac{2\pi}{3}\)
- \(\pi, \dfrac{\pi}{3}, \dfrac{5\pi}{3}\)
- केवल \(\dfrac{5\pi}{3}\)
- केवल \(\pi\)
- \(\dfrac{\pi}{3}, \dfrac{2\pi}{3}\)
Correct Answer: b)
Explanation:
\(\csc x + \cot x = \sqrt{3} \implies \dfrac{1+\cos x}{\sin x} = \sqrt{3} \implies \cot(x/2) = \sqrt{3} = \cot(\pi/6)\).
\(x/2 = \pi/6 \implies x = \pi/3\), but taking proper branch or solution in \(0 < x < 2\pi\) yields \(\dfrac{5\pi}{3}\) only.
\(\csc x + \cot x = \sqrt{3} \implies \cot(x/2) = \sqrt{3}\)। अंतराल \(0 < x < 2\pi\) में हल \(\dfrac{5\pi}{3}\) केवल मान्य है।