[tex]\displaystyle \\
3\tan(2x+15^o)=4 ~~~ \text{ for }~~~ 0^o \leq x \leq 180^o \\ \\
\tan(2x+15^o)= \frac{4}{3} \\ \\
(2x+15^o) = \arctan\Big(\frac{4}{3}\Big) = 53,13010^o ~\approx ~53^o \\ \\
2x+15^o = 53^o \\
2x = 53^o - 15^o \\
2x = 38^o \\ \\
x = \frac{38^o}{2} \\ \\
\boxed{x = 19^o} [/tex]
