Given the equation \(\sin 3x + 2 \cos 3x = 0\).

We can rewrite this as \(\tan 3x = -2\) by dividing both sides by \(\cos 3x\).

Solving \(\tan 3x = -2\), we find the principal value:

\(3x = \tan^{-1}(-2) \approx -63.4^\circ\).

Since \(\tan\) is periodic with period \(180^\circ\), the general solution for \(3x\) is:

\(3x = -63.4^\circ + 180^\circ n\), where \(n\) is an integer.

Solving for \(x\), we have:

\(x = \frac{-63.4^\circ + 180^\circ n}{3}\).

We need \(0^\circ \leq x \leq 180^\circ\), so:

\(0^\circ \leq \frac{-63.4^\circ + 180^\circ n}{3} \leq 180^\circ\).

Solving for \(n\), we find the valid values:

For \(n = 1\), \(x = \frac{116.6^\circ}{3} = 38.9^\circ\).

For \(n = 2\), \(x = \frac{296.6^\circ}{3} = 98.9^\circ\).

For \(n = 3\), \(x = \frac{476.6^\circ}{3} = 158.9^\circ\).

Thus, the solutions are \(x = 38.9^\circ, 98.9^\circ, 158.9^\circ\).