sqrt3sintheta+costheta

প্রশ্ন: \(\sqrt{3}\sin\theta + \cos\theta = 2\)

সমাধান:

বামপক্ষ: \(\sqrt{3}\sin\theta + \cos\theta\)

আমরা বামপক্ষকে \(2\) দিয়ে ভাগ ও গুণ করি:

\(2 \left( \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta \right)\)

আমরা জানি, \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\) এবং \(\sin\frac{\pi}{6} = \frac{1}{2}\)

তাহলে, \(2 \left( \cos\frac{\pi}{6}\sin\theta + \sin\frac{\pi}{6}\cos\theta \right)\)

আমরা জানি, \(\sin(A+B) = \sin A \cos B + \cos A \sin B\)

সুতরাং, \(2 \sin\left(\theta + \frac{\pi}{6}\right)\)

প্রশ্নানুসারে, \(2 \sin\left(\theta + \frac{\pi}{6}\right) = 2\)

অতএব, \(\sin\left(\theta + \frac{\pi}{6}\right) = 1\)

আমরা জানি, \(\sin\frac{\pi}{2} = 1\)

সুতরাং, \(\theta + \frac{\pi}{6} = \frac{\pi}{2}\)

\(\theta = \frac{\pi}{2} - \frac{\pi}{6}\)

\(\theta = \frac{3\pi - \pi}{6}\)

\(\theta = \frac{2\pi}{6}\)

\(\theta = \frac{\pi}{3}\) 🤩

সুতরাং, \(\theta = \frac{\pi}{3}\) হলে \(\sqrt{3}\sin\theta + \cos\theta = 2\) হবে।

সুতরাং, উত্তর: \(\theta = \frac{\pi}{3}\) 🥳