Determine the values of $\theta$, where $0 \leq \theta \leq 360$, such that $\tan \theta=-\frac{1}{\sqrt{3}}$. Place numerical value only of theta in the two blanks with the smaller one first.

Step 1 ：The tangent function has a period of 180 degrees. This means that if \(\tan \theta = -\frac{1}{\sqrt{3}}\), then \(\tan (\theta + 180) = -\frac{1}{\sqrt{3}}\) as well.

Step 2 ：We know that \(\tan 30 = \frac{1}{\sqrt{3}}\) and \(\tan 150 = -\frac{1}{\sqrt{3}}\).

Step 3 ：Therefore, the solutions to the equation \(\tan \theta = -\frac{1}{\sqrt{3}}\) in the interval \(0 \leq \theta \leq 360\) are \(\theta = 150\) and \(\theta = 150 + 180 = 330\).

Step 4 ：Final Answer: The values of \(\theta\) that satisfy the equation \(\tan \theta = -\frac{1}{\sqrt{3}}\) in the interval \(0 \leq \theta \leq 360\) are \(\boxed{150}\) and \(\boxed{330}\).