# Thread: Arctangents in a half-open interval?

1. ## Arctangents in a half-open interval?

The equation is tan^2(X) = 2tan(X) where X is an element of [0, 360)

As you can see, the interval is half-open. My question is, will the solution look like this:

X = arctan A = tan^-1(A) + 180n because inverse tangents come in pairs a half revolution apart

or this:

X = arctan A = tan^-1(A) + 90n because the system is only half open?

(X=Theta but I don't know how to make the symbol)

2. Originally Posted by lauriestar
The equation is tan^2(X) = 2tan(X) where X is an element of [0, 360)

As you can see, the interval is half-open. My question is, will the solution look like this:

X = arctan A = tan^-1(A) + 180n because inverse tangents come in pairs a half revolution apart

or this:

X = arctan A = tan^-1(A) + 90n because the system is only half open?
the interval $\displaystyle 0 \leq x < 360$ just excludes a single value, 360 degrees, nothing more.

$\displaystyle \tan^2{x} = 2\tan{x}$

$\displaystyle \tan^2{x} - 2\tan{x} = 0$

$\displaystyle \tan{x}(\tan{x} - 2) = 0$

$\displaystyle \tan{x} = 0$ ... $\displaystyle x = 0$, $\displaystyle x = 180$ (note that x = 360 is excluded)

$\displaystyle \tan{x} = 2$ ... $\displaystyle x = \arctan(2)$ , $\displaystyle x = \arctan(2) + 180$