Solve the given equation. (Enter your answers as a comma-separated list. Let $$ k $$ be any integer.) $$ \tan (\theta)=-\frac{\sqrt{3}}{3} $$ $$ \theta= $$ $$ \square $$ rad

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The equation is

$$
\tan(\theta)=-\frac{\sqrt{3}}{3}.
$$

Since

$$
\tan\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{3},
$$

we see that the reference angle is $$\frac{\pi}{6}$$.

Because the tangent function is negative, $$\theta$$ must be in the second or fourth quadrants.

- In the second quadrant, the angle is found by using 
  $$
  \theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}.
  $$

- In the fourth quadrant, the angle is given by 
  $$
  \theta=-\frac{\pi}{6}.
  $$

Since the tangent function has a period of $$\pi$$, the general solutions are:

$$
\theta=-\frac{\pi}{6}+k\pi, \quad \theta=\frac{5\pi}{6}+k\pi, \quad \text{where } k \text{ is any integer}.
$$

Thus, the solutions are:

$$
\boxed{-\frac{\pi}{6}+k\pi,\ \frac{5\pi}{6}+k\pi}.
$$
The solutions are $$ \theta = -\frac{\pi}{6} + k\pi $$ and $$ \theta = \frac{5\pi}{6} + k\pi $$, where $$ k $$ is any integer.

Solve the given equation. (Enter your answers as a comma-separated list. Let be any integer.)

rad

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Solve the given equation. (Enter your answers as a comma-separated list. Let be any integer.) rad