Find polar coordinates $(r, \theta)$ of the point, where $r > 0$ and $0 \leq \theta < 2\pi$ Given these Cartesian coordinates: $(2,-3)$
This is my fourth problem of this type, I solved the other 3, but this one has weird numbers and I don't know what to do.
$$\tan\theta = -\frac{3}{2}$$
what would $\theta$ be? The number isn't convenient, if that makes sense. I don't know how to calculate it. I tried entering that into google and it just said it was $.98$ radians. That didn't work as the answer.
As for finding $r$, since I need $(r,\theta)$
I did 
$$x^2 + y^2 = r^2\implies 4 + 9 = r^2\Longrightarrow r^2=13\Longrightarrow r=\sqrt{13}$$
I'm pretty sure that's right but I can't check it because I can't find the accompanying theta value to submit the answer.
 A: Going to post another answer for others' reference since this bubbled up from a while ago, in case anyone has a similar question.
I believe the asker of this question was wondering why that -0.98... does not work, and in fact there are two reasons, I'd believe: one is that -0.98 is not with the range $[0, \tau)$ ($\tau = 2\pi$, I prefer this newfangled notation better), the other is that -0.98 is not an exact answer, and this wants exact answers and I note the other answers here did not seem to address this, thus why I'm posting a new one.
Some seem to operate under the notion that a decimal somehow more "truly" represents the number in question, while an expression like
$$\mathrm{tan}^{-1}\left(-\frac{2}{3}\right)$$
does not. This is not correct. Decimals are only one of many, many possible, and valid, ways of representing numbers, and moreover, given that this number cannot be represented exactly with a decimal that you can either write down or which has a simple pattern to its digits so as to be able to be of easy description even though it cannot be written down in entirety due to its infinite length, they are in fact not useful for questions like these where that, given how the other answers in the author's picture, i.e. https://i.gyazo.com/b1405ca1db16cdc7024180689983b65e.png, were given as exact answers, will not work here. Just because it doesn't have $\pi$ in it, or has a transcendental function, does not mean it is not exact. The transcendental function has a specific definition and gives a specific number to the above (moreover, its range is $\left(-\pi, \pi\right)$ by convention, just as that for $\sqrt{}$ is the nonnegative reals - this convention is part of how the specific symbol is defined, so there is not a question of branch in simply writing this symbol down although there is a question of branch insofar as the original question is stated). Thus it is just as valid as the rest.
Moreover, since that $\theta$ is to be in the range $[0, \tau)$, the way to get that is to note that each non-principal arctangent branch is just $n\tau$ plus the principal branch, so adding $\tau$ to the given arctangent will produce another valid branch, thus the answer is, in conventional notation where $\tau = 2\pi$:
$$\theta = 2\pi + \mathrm{tan}^{-1}\left(-\frac{2}{3}\right)$$
as due to being negative, and no less than $-\pi$, this ends up in the right range. The full polar coordinates of the point are
$$\left(\sqrt{13},\ 2\pi + \mathrm{\tan}^{-1}\left(-\frac{2}{3}\right)\right)$$
and this is a totally viable expression to use with nothing wrong at all.
A: When working with polar coordinates you must use two relations:
$$r = \sqrt{x^2 + y^2}$$
$$\theta = \arctan(y/x)$$
In the case on hand, these relations give us the point: $$(r,\theta)=(\sqrt{13}, -56.3º)$$ 
You can check its correctness by the relation: $$(x,y)=(r*\cos(\theta), r*\sin(\theta))$$ 
You can always calculate the $arctan(y/x)$ without the sign and add the necessary quantity to displace the angle to the quadrant it's placed on; in our case, you just need to change the sign of the angle in question.  
Hope this helps.
A: You correctly found the value of $r$.  It is also true that $\tan\theta = -3/2$. One possible value of $\theta$ is 
$$\theta = \arctan\left(-\frac{3}{2}\right)$$
However,
$$\arctan x: (-\infty, \infty) \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ 
and 
$$-\frac{3}{2} < 0 \implies \arctan\left(-\frac{3}{2}\right) \in \left(-\frac{\pi}{2}, 0\right)$$ 
so simply taking the arctangent of $-3/2$ does not yield a value for $\theta$ in the desired interval.  To find $\theta \in [0, 2\pi)$ such that $\tan\theta = -3/2$, we can use the periodicity of the tangent function.  Since $f(x) = \tan x$ has period $\pi$, the general solution to the equation $\tan\theta = -3/2$ is 
$$\theta = \arctan\left(-\frac{3}{2}\right) + n\pi, n \in \mathbb{Z}$$
What you need to do is determine the values of $n \in \mathbb{Z}$ such that 
$$\theta = \arctan\left(-\frac{3}{2}\right) + n\pi \in [0, 2\pi)$$
A: The two angles are $\tan^{-1} 1.5 = -56.3, 180 -56.3 $ plus $2 k \pi$
