The point $P(t)=(x, y)$ is the terminal point on the unit circle. If $y=-\frac{\sqrt{3}}{2}$ and $x<0$, find the value of cos $t$. \[ \cos \mathrm{t}= \] (Simplify your answer. Type an exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression.

Step 1 ：The point $P(t)=(x, y)$ is on the unit circle, so $x=\cos t$ and $y=\sin t$. We are given that $y=-\frac{\sqrt{3}}{2}$ and $x<0$. Since $x=\cos t$, we need to find the value of $\cos t$ that satisfies these conditions.

Step 2 ：We know that $\cos t$ is negative in the second and third quadrants of the unit circle. Since $y=-\frac{\sqrt{3}}{2}$, we are in the third or fourth quadrant. But since $x<0$, we must be in the third quadrant.

Step 3 ：In the third quadrant, $\cos t$ and $\sin t$ have the same absolute value but opposite signs. Therefore, $\cos t = -\sin t = \frac{\sqrt{3}}{2}$.

Step 4 ：Final Answer: The value of $\cos t$ is $\boxed{-\frac{\sqrt{3}}{2}}$.