2. Determine the general solution of \( 2 \sin 3 x=-\cos 3 x \)

To solve the equation \( 2 \sin 3x = -\cos 3x \), start by rearranging it to: \[ 2 \sin 3x + \cos 3x = 0. \] This can be interpreted as: \[ \cos 3x = -2 \sin 3x. \] Dividing both sides by \( \cos 3x \) (where \( \cos 3x \neq 0 \)), we can use the tangent function: \[ \tan 3x = -\frac{1}{2}. \] Now, the general solution for \( 3x \) is: \[ 3x = \tan^{-1}(-\frac{1}{2}) + n\pi, \quad n \in \mathbb{Z}. \] Next, divide by 3 to solve for \( x \): \[ x = \frac{1}{3} \tan^{-1}(-\frac{1}{2}) + \frac{n\pi}{3}, \quad n \in \mathbb{Z}. \] This gives the general solution for \( x \).