ContourPlot[
{Tan[α Sqrt[β^2 - 1]] == (0.2 Sqrt[1 - k^2 β^2])/Sqrt[β^2 - 1], k = 0.75},
{α, 0, 1.4}, {β, 0, 17}, PlotRange -> {Automatic, {1, 1.5}},
FrameLabel -> Automatic, BaseStyle -> 14]
Например,
k = 0.75;
sol = FullSimplify[NSolve[
Tan[α Sqrt[β^2 - 1]] == (0.2 Sqrt[1 - k^2 β^2])/Sqrt[β^2 - 1], α]];
Когда β равно 1,25
sol /. β -> 1.25
{{α -> 0.1233747751953911}}
Построение с решением sol
expr = sol[[1, 1, 2]];
out = Cases[Table[{expr, β}, {β, 1, 1.5, 0.001}], {_Real, _}];
ListPlot[out, Frame -> True, FrameLabel -> {"α", "β"}, BaseStyle -> 14]