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AI-তৈরি নমুনা উত্তর

প্রদত্ত সমীকরণ:
$\log_4(\sin \theta - \cos 2\theta)^{3\pi} = 2 \left\{ \tan^{-1} 6 + \tan^{-1} \frac{7}{5} \right\}$

ডানপক্ষ হিসাব করি:
যেহেতু $6 > 0$, $\frac{7}{5} > 0$ এবং $6 \times \frac{7}{5} = \frac{42}{5} > 1$,
$\tan^{-1} 6 + \tan^{-1} \frac{7}{5} = \pi + \tan^{-1}\left( \frac{6 + \frac{7}{5}}{1 - 6 \times \frac{7}{5}} \right)$
$= \pi + \tan^{-1}\left( \frac{\frac{37}{5}}{-\frac{37}{5}} \right) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

সুতরাং ডানপক্ষ $= 2 \times \frac{3\pi}{4} = \frac{3\pi}{2}$

এখন বামপক্ষ:
$\log_4(\sin \theta - \cos 2\theta)^{3\pi} = 3\pi \log_4(\sin \theta - \cos 2\theta)$

অতএব,
$3\pi \log_4(\sin \theta - \cos 2\theta) = \frac{3\pi}{2}$
$\implies \log_4(\sin \theta - \cos 2\theta) = \frac{1}{2}$
$\implies \sin \theta - \cos 2\theta = 4^{1/2} = 2$
$\implies \sin \theta - (1 - 2\sin^2 \theta) = 2$
$\implies 2\sin^2 \theta + \sin \theta - 3 = 0$
$\implies 2\sin^2 \theta + 3\sin \theta - 2\sin \theta - 3 = 0$
$\implies \sin \theta (2\sin \theta + 3) - 1(2\sin \theta + 3) = 0$
$\implies (\sin \theta - 1)(2\sin \theta + 3) = 0$

যেহেতু $-1 \le \sin \theta \le 1$, তাই $2\sin \theta + 3 \neq 0$।
$\therefore \sin \theta - 1 = 0 \implies \sin \theta = 1$

সীমা: $-2\pi \le \theta \le 2\pi$
$\sin \theta = 1$ হলে,
$\theta = \frac{\pi}{2}, \frac{\pi}{2} - 2\pi = -\frac{3\pi}{2}$

নির্ণেয় সমাধান: $\theta = -\frac{3\pi}{2}, \frac{\pi}{2}$