MHTCET 13 Apr 2026 Shift 1 Paper

Concept:Convert $\tan 5x$ into $\frac{\sin 5x}{\cos 5x}$ and use product-to-sum identities to simplify the equation.Explanation:Given equation:$$\cos 3x \cdot \tan 5x = \sin 7x$$Since $\tan 5x = \frac{\sin 5x}{\cos 5x}$, we need $\cos 5x \neq 0$.Multiplying both sides by $\cos 5x$:$$\cos 3x \sin 5x = \sin 7x \cos 5x$$Using the identity $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$:$$\cos 3x \sin 5x = \frac{1}{2}[\sin 8x + \sin 2x]$$$$\sin 7x \cos 5x = \frac{1}{2}[\sin 12x + \sin 2x]$$Equating and simplifying:$$\sin 8x = \sin 12x$$$$\sin 8x - \sin 12x = 0$$Using $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$:$$2\cos 10x \sin(-2x) = 0$$Therefore, $\sin 2x = 0$ or $\cos 10x = 0$.For $x \in \left[0, \frac{\pi}{2}\right)$:$\sin 2x = 0$ gives $x = 0$ → 1 solution.$\cos 10x = 0$ gives:$$10x = \frac{(2k+1)\pi}{2} \Rightarrow x = \frac{(2k+1)\pi}{20}$$Since $0 \le x < \frac{\pi}{2}$, the possible values are:$$x = \frac{\pi}{20}, \frac{3\pi}{20}, \frac{5\pi}{20}, \frac{7\pi}{20}, \frac{9\pi}{20}$$This gives 5 solutions.None of these values make $\cos 5x = 0$, so all are valid.Total number of solutions:$$1 + 5 = 6$$Answer:$6$Option D.