Write the complex number 3-i in trigonometric (polar) form, with theta in the interval [0 degrees, 360 degrees). Round to the nearest tenth.
You want:
$\displaystyle 3-i=r\,\text{cis}(\theta)$
From your other topic where we went from polar to rectangular form, you should be able to see that:
(1) $\displaystyle r\cos(\theta)=3$
(2) $\displaystyle r\sin(\theta)=-1$
and by Pythagoras we know:
(3) $\displaystyle r^2\sin^2(\theta)+r^2\cos^2(\theta)=r^2$
What do we get if we divide (2) by (1)? What quadrant is the angle $\displaystyle \theta$ in?
What do we get if we use (1) and (2) to substitute into (3)?
What I mean by dividing (2) by (1) is you should get:
$\displaystyle \frac{r\sin(\theta)}{r\cos(\theta)}=\frac{-1}{3}$
which simplifies to:
$\displaystyle \tan(\theta)=-\frac{1}{3}$
Now, when we use the inverse tangent function, we need to be mindful of which quadrant the angle is actually in. Which quadrant is the sine function negative and the cosine function positive for a given angle? Your calculator is going to give you:
$\displaystyle -90^{\circ}<\theta<90^{\circ}$
and care must be taken to use this, and the fact that $\displaystyle \tan\left(\theta+k\cdot180^{\circ} \right)=\tan(\theta)$ where $\displaystyle k\in\mathbb{Z}$ ($\displaystyle k$ is an integer) to put the angle in the desired form.
So first, what does your calculator tell you the angle is?