Trigonometry — Hard JEE math MCQ
The general solution of $\sin 2x = \cos 3x$ is:
- A. $x = n\pi + \frac{\pi}{10}$ or $x = n\pi - \frac{\pi}{2}$
- B. $x = n\pi + (-1)^n \frac{\pi}{10}$ or $x = 2n\pi \pm \frac{\pi}{2}$
- C. $x = \frac{n\pi}{5} + \frac{\pi}{10}$ or $x = n\pi - \frac{\pi}{2}$
- D. $x = \frac{(4n+1)\pi}{10}$ or $x = (2n-1)\frac{\pi}{2}$
Solution
$\sin 2x = \cos 3x$
Using $\sin \theta = \cos\left(\frac{\pi}{2} - \theta\right)$:
$$\cos\left(\frac{\pi}{2} - 2x\right) = \cos 3x$$
This gives: $\frac{\pi}{2} - 2x = 2n\pi \pm 3x$
**Case 1:** $\frac{\pi}{2} - 2x = 2n\pi + 3x$
$$\frac{\pi}{2} = 2n\pi + 5x$$
$$x = \frac{\pi - 4n\pi}{10} = \frac{(1-4n)\pi}{10} = \frac{(4n+1)\pi}{10}$$ (replacing $n$ with $-n$)
**Case 2:** $\frac{\pi}{2} - 2x = 2n\pi - 3x$
$$\frac{\pi}{2} + x = 2n\pi$$
$$x = 2n\pi - \frac{\pi}{2} = (2n-1)\frac{\pi}{2}$$