Question

Solve 3tan(2x+15°)=4 for 0°≤x≤180°

Answer 1

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