AI-তৈরি নমুনা উত্তর
দেওয়া আছে,3 sin 2 x − 1 cos 2 x = 4 \frac{\sqrt{3}}{\sin 2x} - \frac{1}{\cos 2x} = 4 s i n 2 x 3 − c o s 2 x 1 = 4
বা, 3 cos 2 x − sin 2 x sin 2 x cos 2 x = 4 \frac{\sqrt{3}\cos 2x - \sin 2x}{\sin 2x \cos 2x} = 4 s i n 2 x c o s 2 x 3 c o s 2 x − s i n 2 x = 4 3 cos 2 x − sin 2 x = 4 sin 2 x cos 2 x \sqrt{3}\cos 2x - \sin 2x = 4\sin 2x \cos 2x 3 cos 2 x − sin 2 x = 4 sin 2 x cos 2 x 3 cos 2 x − sin 2 x = 2 ( 2 sin 2 x cos 2 x ) = 2 sin 4 x \sqrt{3}\cos 2x - \sin 2x = 2(2\sin 2x \cos 2x) = 2\sin 4x 3 cos 2 x − sin 2 x = 2 ( 2 sin 2 x cos 2 x ) = 2 sin 4 x
উভয়পক্ষকে 2 2 2 3 2 cos 2 x − 1 2 sin 2 x = sin 4 x \frac{\sqrt{3}}{2}\cos 2x - \frac{1}{2}\sin 2x = \sin 4x 2 3 cos 2 x − 2 1 sin 2 x = sin 4 x sin π 3 cos 2 x − cos π 3 sin 2 x = sin 4 x \sin\frac{\pi}{3}\cos 2x - \cos\frac{\pi}{3}\sin 2x = \sin 4x sin 3 π cos 2 x − cos 3 π sin 2 x = sin 4 x sin ( π 3 − 2 x ) = sin 4 x \sin\left(\frac{\pi}{3} - 2x\right) = \sin 4x sin ( 3 π − 2 x ) = sin 4 x sin 4 x − sin ( π 3 − 2 x ) = 0 \sin 4x - \sin\left(\frac{\pi}{3} - 2x\right) = 0 sin 4 x − sin ( 3 π − 2 x ) = 0 2 cos ( 4 x + π 3 − 2 x 2 ) sin ( 4 x − π 3 + 2 x 2 ) = 0 2\cos\left(\frac{4x + \frac{\pi}{3} - 2x}{2}\right) \sin\left(\frac{4x - \frac{\pi}{3} + 2x}{2}\right) = 0 2 cos ( 2 4 x + 3 π − 2 x ) sin ( 2 4 x − 3 π + 2 x ) = 0 2 cos ( x + π 6 ) sin ( 3 x − π 6 ) = 0 2\cos\left(x + \frac{\pi}{6}\right) \sin\left(3x - \frac{\pi}{6}\right) = 0 2 cos ( x + 6 π ) sin ( 3 x − 6 π ) = 0
সুতরাং,cos ( x + π 6 ) = 0 \cos\left(x + \frac{\pi}{6}\right) = 0 cos ( x + 6 π ) = 0 ⟹ x + π 6 = ( 2 n + 1 ) π 2 \implies x + \frac{\pi}{6} = (2n + 1)\frac{\pi}{2} ⟹ x + 6 π = ( 2 n + 1 ) 2 π ⟹ x = n π + π 2 − π 6 \implies x = n\pi + \frac{\pi}{2} - \frac{\pi}{6} ⟹ x = nπ + 2 π − 6 π ⟹ x = n π + π 3 , n ∈ Z \implies x = n\pi + \frac{\pi}{3}, \quad n \in \mathbb{Z} ⟹ x = nπ + 3 π , n ∈ Z
অথবা, sin ( 3 x − π 6 ) = 0 \sin\left(3x - \frac{\pi}{6}\right) = 0 sin ( 3 x − 6 π ) = 0 ⟹ 3 x − π 6 = m π \implies 3x - \frac{\pi}{6} = m\pi ⟹ 3 x − 6 π = mπ ⟹ 3 x = m π + π 6 \implies 3x = m\pi + \frac{\pi}{6} ⟹ 3 x = mπ + 6 π ⟹ x = m π 3 + π 18 , m ∈ Z \implies x = \frac{m\pi}{3} + \frac{\pi}{18}, \quad m \in \mathbb{Z} ⟹ x = 3 mπ + 18 π , m ∈ Z
যেখানে sin 2 x ≠ 0 \sin 2x \neq 0 sin 2 x = 0 cos 2 x ≠ 0 \cos 2x \neq 0 cos 2 x = 0
নির্ণেয় সমাধান: x = n π + π 3 x = n\pi + \frac{\pi}{3} x = nπ + 3 π x = m π 3 + π 18 x = \frac{m\pi}{3} + \frac{\pi}{18} x = 3 mπ + 18 π n , m ∈ Z n, m \in \mathbb{Z} n , m ∈ Z