Start with the equation \(3 \tan(2x + 15^\circ) = 4\).

Divide both sides by 3 to isolate the tangent function:

\(\tan(2x + 15^\circ) = \frac{4}{3}\).

Find the angle whose tangent is \(\frac{4}{3}\):

\(2x + 15^\circ = \tan^{-1}\left(\frac{4}{3}\right)\).

Using a calculator, \(\tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ\).

Since tangent has a period of \(180^\circ\), the general solution is:

\(2x + 15^\circ = 53.13^\circ + 180^\circ k\) or \(2x + 15^\circ = 233.13^\circ + 180^\circ k\), where \(k\) is an integer.

For \(k = 0\), solve for \(x\):

\(2x + 15^\circ = 53.13^\circ\)

\(2x = 53.13^\circ - 15^\circ\)

\(2x = 38.13^\circ\)

\(x = 19.065^\circ\)

Rounding to one decimal place, \(x = 19.1^\circ\).

For the second solution:

\(2x + 15^\circ = 233.13^\circ\)

\(2x = 233.13^\circ - 15^\circ\)

\(2x = 218.13^\circ\)

\(x = 109.065^\circ\)

Rounding to one decimal place, \(x = 109.1^\circ\).

Thus, the solutions are \(x = 19.1^\circ\) or \(x = 109.1^\circ\).