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Elis [28]
3 years ago
12

Solve 3tan(2x+15°)=4 for 0°≤x≤180°

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
5 0
   
\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}



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