Solve for $t, 0 \leq t<2 \pi$. \[ \begin{array}{l} 24 \sin (t) \cos (t)=-9 \cos (t) \\ t= \end{array} \]

Step 1 ：The given equation is a trigonometric equation. To solve for $t$, we can first simplify the equation by dividing both sides by $\cos(t)$, assuming $\cos(t) \neq 0$. This will give us $24 \sin(t) = -9$.

Step 2 ：We can then solve for $t$ by taking the arcsine of both sides. However, we need to consider the range of $t$ which is $0 \leq t < 2 \pi$. This means we need to consider the unit circle and the fact that sine is negative in the third and fourth quadrants.

Step 3 ：The solution obtained is approximately $3.53$. However, we need to consider that the sine function is negative in the third and fourth quadrants. Therefore, we need to find the equivalent angle in the third or fourth quadrant. We can do this by subtracting the obtained solution from $\pi$ and $2\pi$ respectively.

Step 4 ：The equivalent angles in the third and fourth quadrants are approximately $2.76$ and $2.76$ respectively. However, these angles are not in the range $0 \leq t < 2 \pi$. Therefore, the original solution $t = 3.53$ is the only solution in the given range.

Step 5 ：Final Answer: The solution to the equation $24 \sin (t) \cos (t)=-9 \cos (t)$ in the range $0 \leq t < 2 \pi$ is $t = \boxed{3.53}$.