#### Answer

$6(\cos(240^{\circ})+i\sin(240^{\circ}))$

#### Work Step by Step

To find the trigonometric form of a complex number from its Cartesian form, two things must be done. Firstly, the modulus of the complex number must be found. Secondly, the argument of the complex number must be found.
To find the modulus of the complex number:
Apply Pythagoras theorem to the coefficients of the complex number. i.e. $r=\sqrt{(-3)^{2}+(-3\sqrt{3})^{2}}\\=\sqrt{36}\\=6$
To find the argument of the complex number:
First, find the basic argument of the complex number this is done by finding $\arctan$ of the fraction of the absolute value of the imaginary number coefficient over the coefficient of the real number. Therefore, the basic argument is $\arctan({\frac{3\sqrt{3}}{3}})\\=\arctan(\sqrt{3})\\=60^{\circ}$
Second, identify the quadrant that the line graph of the complex number lie in. Below are the conditions for the four quadrants that the complex number can lie in.
First quadrant: both coefficient of real and imaginary parts are positive [argument=basic argument]
Second quadrant: coefficient of real part is negative while coefficient of imaginary part is positive [argument=$180^{\circ}$-basic argument]
Third quadrant: coefficient of real and imaginary part is negative [argument=$180^{\circ}$+basic argument]
Fourth quadrant: coefficient of real part is positive while coefficient of imaginary part is negative [argument=$360^{\circ}$-basic argument]
Since both the coefficient of the real part and imaginary part is negative, the Cartesian equation lies in the third quadrant and the argument for the complex number trigonometric form is thus $180^{\circ}+60^{\circ}=240^{\circ}$
Thus, the trigonometric form of the Cartesian equation of the complex number $-3-3i\sqrt{3}$ is $6(\cos(240^{\circ})+i\sin(240^{\circ}))$