square root of 3 times the cot times 3x plus one equals 0..
The cotangent of what? Do you mean $\displaystyle \sqrt3\cot\left(3x + 1\right) = 0$? Or maybe $\displaystyle \sqrt{3\cot\left(3x + 1\right)} = 0$? Or $\displaystyle \sqrt3\cot\left(3x\right) + 1 = 0$? $\displaystyle \sqrt{3\cot\left(3x\right)} + 1 = 0$? $\displaystyle \sqrt{3\cot\left(3x\right) + 1} = 0$? $\displaystyle \sqrt3\left[\cot\left(3x\right) + 1\right] = 0$?
Is this what you mean?
$\displaystyle \sqrt{3}\cot(3x)+1=0$
Is there a restriction on x? (i.e. $\displaystyle 0 \leq x <2\pi$)?
After some modifications, we get:
$\displaystyle \cot(3x)=-\frac{1}{\sqrt{3}}$
Note that $\displaystyle \cot\theta=-\frac{1}{\sqrt{3}}$ when $\displaystyle \theta=\frac{2\pi}{3}\text{,} \ \frac{5\pi}{3}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}$
Thus, $\displaystyle 3x=\frac{2\pi}{3}\text{,} \ \frac{5\pi}{3}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}$
$\displaystyle \implies x=\frac{2\pi}{9}\text{,} \ \frac{5\pi}{9}\text{,} \ \dots \ \frac{(3n-1)\pi}{3}; \ n\in\mathbb{N}$.
I hope this helped (assuming that I interpreted the question properly)!
Ok then! Referring back to my solution (I had it in radians, now I'm gonna convert it to degrees...), we saw that:
$\displaystyle x=\frac{2(360^o)}{9}\text{,} \ \frac{5(360^o)}{9}\text{,} \ \frac{8(360^o)}{9}\text{,} \ \frac{11(360^o)}{9}\text{,} \ \frac{14(360^o)}{9}\text{,} \ \dots \ \frac{(3n-1)(360^o)}{9}; \ n\in\mathbb{N}$
After simplifying we get:
$\displaystyle \color{red}\boxed{x=80^o\text{,} \ 200^o\text{,} \ 320^o}$
The ones after 320 are outside of the domain ($\displaystyle 0\leq x<360^o$).
Hope this made sense!
Forgive me, but I made a "little" error
$\displaystyle \pi = 180^o\neq 360^o$
Thus, $\displaystyle \color{red}\boxed{x=40^o\text{,} \ 100^o\text{,} \ 160^o\text{,} \ 220^o\text{,} \ 280^o\text{,} \ 340^o}$
To get to the new answers, divide the old ones by 2 (you will have more answers...I hope you can follow).
Sorry about that!