Question

Find the general solution of the following equation:

\[\sin 2x = \cos 3x\]

Answer 1

We have:

\[\sin2x = \cos3x\]

\[\cos3x = \sin2x\]

⇒ \[\cos3x = \cos \left( \frac{\pi}{2} - 2x \right)\]

⇒ \[3x = 2n\pi \pm \left( \frac{\pi}{2} - 2x \right), n \in Z\]

On taking positive sign, we have:
\[3x = 2n\pi + \left( \frac{\pi}{2} - 2x \right)\]

⇒ \[5x = 2n\pi + \frac{\pi}{2}\]

⇒ \[x = \frac{2n\pi}{5} + \frac{\pi}{10}\]

⇒ \[x = (4n + 1)\frac{\pi}{10}\]

\[n \in Z\]

 Now, on taking negative sign, we have:

\[3x = 2n\pi - \frac{\pi}{2} + 2x, n \in Z\]

⇒ \[x = 2n\pi - \frac{\pi}{2}\]

⇒ \[x = (4n - 1)\frac{\pi}{2}, n \in Z\]