xemynort
xemynort
10-10-2016
Mathematics
contestada
solve 3tan(2x+15°)=4 for 0°≤x≤180°
Respuesta :
Tucon
Tucon
10-10-2016
[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]
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