HIV inflicts its damage by infecting healthy CD4 + T cells (a type of white blood cell) that are necessary to fight infection. As the virus embeds in a T cell and the immune system produces more of these cells to fight the infection, the virus propagates in an opportunistic fashion. Normally T cells are produced at a rate $s$ and die at a rate $d$. The HIV virus is present in the bloodstream in the infected individual. These viruses in the bloodstream, called free viruses, infect healthy T cells at a rate $\beta$. Also, the viruses reproduce through the T cell multiplication process or otherwise at a rate $k$. Free viruses die at a rate c. Infected T cells die at a rate $\mu$.
A simple mathematical model to describe how the HIV/AIDS virus infects healthy cells is given by the following equations:
$$\begin{align}\frac{dT}{dt} &= s-dT-\beta T v \\ \frac{dT^*}{dt} &= \beta T v - \mu T^* \\ \frac{dv}{dt} &= kT^* -cv\end{align}$$
I am told that the system has two equilibrium points given by
$$(T_0, T_0^*,v_0) = (\frac{s}{d},0,0)$$ and $$(T_0, T_0^*,v_0) = (\frac{c\mu}{\beta k}, \frac{s}{\mu}-\frac{cd}{\beta k}, \frac{sk}{c\mu}-\frac{d}{\beta})$$
I am asked to prove that the equilibrium points are the ones shown above but I am having trouble proving this after setting the derivatives equal to $0$. I can obtain the first equilibirum point easily if $\mu$ is zero but otherwise not. How would I solve this system for the two equilbirium points? My differential equations knowledge is quite rusty.
