5. Draw the tangent line to the curve $y=3x-x^{3}$ through the point $A(2,-2)$. Then the equation of the tangent line is $\qquad$ .
5. $y=-2$ or $9 x+y-16=0$.

Let the point of tangency be $P\left(x_{0}, y_{0}\right)$, then the equation of the tangent line at point $P$ is $\boldsymbol{y}-\boldsymbol{y}_{0}=k\left(x-x_{0}\right)$, where, $k=\left.y^{\prime}\right|_{x=x_{0}}=3-3 x_{0}^{2}$.
Since the tangent line passes through $A(2,-2)$, and point $P$ is on the curve, we have
$$
\left\{\begin{array}{l}
-2-y_{0}=\left(3-3 x_{0}^{2}\right)\left(2-x_{0}\right), \\
y_{0}=3 x_{0}-x_{0}^{3} .
\end{array}\right.
$$

Eliminating $y_{0}$ gives $x_{0}^{3}-3 x_{0}^{2}+4=0$, that is,
$$
\left(x_{0}+1\right)\left(x_{0}^{2}-4 x+4\right)=0 \text {. }
$$

Solving, we get $x_{0}=-1$ or $x_{0}=2$.
When the point of tangency is $P(-1,-2)$, $k=0$, the equation of the tangent line is $y=-2$.

When the point of tangency is $P(2,-2)$ (which is point $A$), $k=-9$, the equation of the tangent line is
$y-(-2)=-9(x-2)$, that is,
$$
9 x+y-16=0 \text {. }
$$

Note: We cannot only find the tangent line passing through point $A$.