Finding required drug concentration given Kdiss The enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9 mol/l). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7 mol/l).
Compute the total concentration of [D1]tot that is needed to bind 90% of the HIV-RT at the given concentration [E]tot. You do NOT have to consider competition betwwen the drugs D1 and D2! They are administered separately.

I understand how to calculate the concentration when [E]eq = [ED]eq when 50% of [D]eq is bound to HIV-RT because Kdiss = [D]eq and it makes for a simple calculation because [E]eq and [ED]eq cancel out when calculating Kdiss, but what changes mathematically need to be made to account for the change in ratio (90% rather than 50%)? Any help would be appreciated.
Chemical reaction is

$$\ce{D1 + E &lt;=&gt; ED1}$$

Initial concentrations: $[\ce{D1}]_\textrm{tot}; \quad [\ce{E}]_\textrm{tot}; \quad 0$.

If $X$ percent of $E$ "reacts", then stoichiometry tells us that:

Concentrations at equilibrium: $[\ce{D1}]_\textrm{tot} (1- X) ;\quad [\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X;  \quad [\ce{D1}]_\textrm{tot} X$.

The equilibrium expression is

$$K_\textrm{diss} = \frac{ [E][D_1]}{[ED_1]}$$

Therefore

$$K_\textrm{diss} = \frac{ [\ce{D1}]_\textrm{tot} (1- X) ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)}{[\ce{D1}]_\textrm{tot} X}$$

Which we can simplify to

$$K_\textrm{diss} = \frac{(1- X)}{X} ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)$$

So you can solve this for $X = 0.90$, $[\ce{E}]_\textrm{tot} = 2.00 \ \mu M$ and $K = 1.00 \ nM$.