গণিত প্রশ্ন

AI-তৈরি নমুনা উত্তর

প্রদত্ত সমীকরণ:
$\sin\theta + \sin 2\theta + \sin 3\theta = 1 + \cos\theta + \cos 2\theta$

$\implies (\sin 3\theta + \sin\theta) + \sin 2\theta = (1 + \cos 2\theta) + \cos\theta$
$\implies 2\sin\left(\frac{3\theta + \theta}{2}\right)\cos\left(\frac{3\theta - \theta}{2}\right) + \sin 2\theta = 2\cos^2\theta + \cos\theta$
$\implies 2\sin 2\theta\cos\theta + \sin 2\theta = 2\cos^2\theta + \cos\theta$
$\implies \sin 2\theta (2\cos\theta + 1) = \cos\theta (2\cos\theta + 1)$
$\implies \sin 2\theta (2\cos\theta + 1) - \cos\theta (2\cos\theta + 1) = 0$
$\implies (2\cos\theta + 1)(\sin 2\theta - \cos\theta) = 0$
$\implies (2\cos\theta + 1)(2\sin\theta\cos\theta - \cos\theta) = 0$
$\implies \cos\theta(2\cos\theta + 1)(2\sin\theta - 1) = 0$

হয়, $\cos\theta = 0$
$\implies \theta = (2n + 1)\frac{\pi}{2}, \quad n \in \mathbb{Z}$

অথবা, $2\cos\theta + 1 = 0$
$\implies \cos\theta = -\frac{1}{2} = \cos\left(\frac{2\pi}{3}\right)$
$\implies \theta = 2n\pi \pm \frac{2\pi}{3}, \quad n \in \mathbb{Z}$

অথবা, $2\sin\theta - 1 = 0$
$\implies \sin\theta = \frac{1}{2} = \sin\left(\frac{\pi}{6}\right)$
$\implies \theta = n\pi + (-1)^n\frac{\pi}{6}, \quad n \in \mathbb{Z}$

নির্ণেয় সমাধান:
$\theta = (2n + 1)\frac{\pi}{2}, \quad 2n\pi \pm \frac{2\pi}{3}, \quad n\pi + (-1)^n\frac{\pi}{6}; \quad n \in \mathbb{Z}$