General solution of trig equations

[ausmathgenius420]

Hi, I was wondering if there is any convention when determining the general solution for trig equations.

The problem is [math]tan(2x)= -\sqrt3[/math]
Using the general solution for tanx=a the next line of working should be:
[math]2x=n\pi + arctan( -\sqrt3)[/math]
My question is what should I set [imath]arctan(-\sqrt3)[/imath] to?.. My textbook uses [imath]- \frac{\pi}{3}[/imath] however it could also be 2pi/3 or 5pi/3 if you only use positive angles. Does it matter which angle I use? - I know they will all give me the same answers but is one technically correct?

 

[JeffM]

Technically, the function [imath]arctan( \theta ) [/imath] is defined only if [imath]-\dfrac{\pi}{2} < \theta < \dfrac{ \pi }{2}.[/imath]

The arctan function is a function. Therefore, by definition, it provides an unambiguous result. That is the reason for your book’s use of [imath]-\sqrt{3}.[/imath]

 

[ausmathgenius420]

Technically, the function [imath]arctan( \theta ) [/imath] is defined only if [imath]-\dfrac{\pi}{2} < \theta < \dfrac{ \pi }{2}.[/imath].

The arctan function is a function. Therefore, by definition, it provides an unambiguous result. That is the reason for your book’s use of [imath]-\sqrt{3}.[/imath]

Thank you!

 

[JeffM]

I realize upon reflection that my answer may be misinterpreted.

If we define the arc tangent function as the inverse of the tangent function, we must do so over a domain such that the tangent function is monotonically increasing. So I should have said.

[math]arctan( \theta ) \equiv tan^{-1} (\theta) \text { where } - \dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}.[/math]
If, however, we apply the language of domain and range to the arctan function as a stand-alone function, its domain is all real numbers, and its range is what is restricted.

Your question, however, was about the solution to trigonometric equations, and there it is the restriction on the domain of the trigonometric functions that is relevant.