How can I solve Tan[t] - t == F[x] for t as a function of x?

How can I solve the equation Tan[t] - t = Ax, where A is a constant for t[x]?

I know that it's impossible to get a close-form solution, but how can I get a numerical approximation with Mathematica?

A supplement to above two wonderful answers:

Notice that any two of the branches of curve $C$ defined by $f(t)=\tan(t)-t$ are identical with only a translation of $\boldsymbol{\mathrm{v}}_n=(n \pi,-n \pi)^{\mathrm T}$ :

$$\left( \begin{array}{c} t \\ \tan(t)-t \\ \end{array} \right)+ \left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)n\pi = \left( \begin{array}{c} t+n\pi \\ \tan(t+n\pi)-(t+n\pi) \\ \end{array} \right)$$

graph1 = Plot[Tan[t] - t, {t, -3 Pi, 3 Pi},
PlotRange -> {-4, 12}, AspectRatio -> Automatic,
MaxRecursion -> Infinity,
PlotStyle -> Lighter[Blue],
Exclusions -> {1/(Tan[t] - t) == 0},
ExclusionsStyle -> Directive[Gray, Dashed]
] // Quiet;

graph2 = Plot[Tan[t] - t, {t, -Pi/2, Pi/2},
PlotRange -> {-4, 12},
MaxRecursion -> Infinity,
PlotStyle -> Directive[Blue, Thick]
] // Quiet;

Manipulate[
solset = t /. FindRoot[Tan[t] - t == F, {t, #}] &
/@ (.9 \[Pi] Range[-2, 2] + .3);
Show[{graph1, graph2},
Frame -> True,
FrameStyle -> Directive[Opacity],
FrameTicks -> {
{Range[-4, 12], Range[-4, 12]},
{Range[-5, 5, 2] Pi/2, Range[-5, 5, 2] Pi/2}},
FrameTicksStyle -> Directive[Opacity, Black, Bold, 20],
Axes -> None,
GridLines -> {None, F + Pi Range[-2, 2]},
GridLinesStyle -> Directive[Lighter[Brown], Dashed],
Epilog -> {
Brown, Thickness[.005],
Line[{{-3 Pi, F}, {3 Pi, F}}],
Darker[Green, .3],
MapIndexed[
Disk[{#1, F} - (#2[] - 3) Pi {1, -1}, .2] &,
solset],
Red,
Disk[{#, F}, .2] & /@ solset,
MapIndexed[
Arrow[{{#1, F} - (#2[] - 3) Pi {1, -1}, {#1, F}}] &,
solset]
},
ImageSize -> 600],
{{F, 4}, -2, 10}] So for any $F$, to get all solutions of $\tan(t)-t=F$ (the red points in above graphics), just find all intersection points of the center branch of $C$ and horizon lines $l_n(t)=F+n \pi$, where $n\in\mathbb{Z}$ (the green points in above graphics), then translate them with proper vectors.

To find those green points, you can adopt Jens' InverseFunction method, or Alexei's approximate analytical solution, or it is also possible to derive an approximate formal series by functions such as InverseSeries:

seriesZero = InverseSeries[Series[Tan[t] - t, {t, 0, 9}], F] seriesInfinity = InverseSeries[Series[Tan[t] - t, {t, Pi/2, 9}], F] Plot[Evaluate[{
Normal@seriesZero,
Normal@seriesInfinity,
InverseFunction[Function[t, Tan[t] - t]][F]
}], {F, 0, 8},
PlotRange -> {0, Pi/2},
PlotStyle -> (ColorData["Rainbow"] /@ {0, .5, 1}),
Frame -> True,
FrameLabel -> (Style[#, Italic, 20, Bold] & /@ {"F", "t"})] So we have a function of $x$ as $$x\mapsto(F=F(x))\mapsto t\text{ .}$$

The most convenient way to answer this question is by using InverseFunction which allows you to hide the details of finding the numerical inverse:

g = InverseFunction[Function[{t}, Tan[t] - t]]

Then your equation Tan[t] - t == a x can be solved by simply saying

g[a x]

You might look for the approximate analytical solution as follows. Let us first denote y=Ax. We will substitute it back later. This is the table of solutions of your equation with 0<=y<=Pi/2:

tb = Table[{y, FindRoot[Tan[t] - t == y, {t, 1.2}][[1, 2]]}, {y,
0, \[Pi]/2 - 0.02, 0.1}];

Now one can fit this list, and get an analytical solution out of it:

Manipulate[
lstPL = ListPlot[tb, PlotStyle -> Red,
AxesLabel -> {Style["y", 14, Italic],
Style["\!$$\*SubscriptBox[\(t$$, $$sol$$]\)", 14, Italic]}];

ft = Fit[tb, {y^a, y^(a + b)}, y];

pl = Plot[ft, {y, 0, \[Pi]/2}];

Show[{lstPL, pl},
Epilog ->
Inset[Row[{Style["t=", 14, Italic],
Style[NumberForm[ft, {3, 2}], 14, Italic]}],
Scaled[{0.7, 0.3}]]],

{{a, 0.306}, 0, 1}, {{b, 0.458}, 0, 1}]

This results in such a fit, see the image: . You get finally $t_{\text{sol}}\approx 1.39 (A x)^{0.31}-0.26 (A x)^{0.76}$