Solve equation 3tan(2x+15)=4 for 0<x<180

[Kayla65]

Solve equation 3tan(2x+15)=4 for 0<x<180

My Workings:
tan(2x+15)=4/3
2x+15=arctan(4/3)
2x=arctan(4/3)-15
x=(arctan(4/3)-15)/2
x=-7.036

tan^1(-7.036)=-81.91
-81.91+180=98.089

Am I correct or did I mess up along the way:?:

 

[Ishuda]

Solve equation 3tan(2x+15)=4 for 0<x<180

My Workings:
tan(2x+15)=4/3
2x+15=arctan(4/3)
2x=arctan(4/3)-15
x=(arctan(4/3)-15)/2
x=-7.036

tan^1(-7.036)=-81.91
-81.91+180=98.089

Am I correct or did I mess up along the way:?:

You are correct up to this point
tan^1(-7.036)=-81.91
-81.91+180=98.089
I'm not sure what you are doing at this point.

If we go back to x=-7.036, that is a proper answer except for the restriction 0<x<180. What we have, going back another step is

x=(arctan(4/3)-15)/2 = \(\displaystyle [0.9273\, +\, n\, \pi\, -\, 15]\, /\, 2\)

You have done n=0. So what n would make x between 0 and 180 degrees [between 0 and 3.14159]?

 

[Steven G]

Solve equation 3tan(2x+15)=4 for 0<x<180

My Workings:
tan(2x+15)=4/3
2x+15=arctan(4/3)
2x=arctan(4/3)-15
x=(arctan(4/3)-15)/2
x=-7.036

tan^1(-7.036)=-81.91
-81.91+180=98.089

Am I correct or did I mess up along the way:?:

Not trying to be rude but of course you are wrong. You are told that x is in between 0 and 180, yet you say that x=-7.036!

 

[Ishuda]

Not trying to be rude but of course you are wrong. You are told that x is in between 0 and 180, yet you say that x=-7.036!

Jomo,

I know, strictly speaking, that you are right but I think that was supposed to be an intermediate step, like being asked to find x, 0<x<180 degrees, when tan(x)=-1. You start with x=-\(\displaystyle \pi/4\)=-45 degrees and add 180 degrees to get 135 degrees which is in the proper interval.

 

[HallsofIvy]

Recheck the statement of the problem. Did it say that x was between 0 and 180 or that 2x+ 15, the argument of the tangent function was between 0 and 180?