# Plotting sections through plane

I have given the following function:

B[x_] := Sqrt[x^2 - (9 - 3 x)^2]


where x is also dependent of several cos-function. All in all I wanna plot the following function:

ContourPlot[
B[1/2 (Cos[Subscript[k, x]] +
2*Cos[Subscript[k, x]/2]*Cos[Sqrt[3] Subscript[k, z]/2] +
6)], {Subscript[k, x], -6 Pi/3, 6 Pi/3}, {Subscript[k,
z], -6 Pi/3, 6 Pi/3}, PlotRange -> Full, ImageSize -> 500,
FrameTicks -> {Pi/3 Range[-6, 6], Pi/3 Range[-6, 6]},
FrameLabel -> {"k_x", "k_z"}, PlotLegends -> Automatic]


The plot is looking like this: But now I wanna plot sections (or cuts) through the plane. I tried to calculate this:

Solve[B[1/
2 (Cos[Subscript[k, x]] +
2*Cos[Subscript[k, x]/2]*Cos[Sqrt[3] Subscript[k, z]/2] + 6)] == Pi/2, {Subscript[k, x], Subscript[k, z]}]


Plot[-((2 ArcCos[
1/8 (7 Cos[Subscript[k, x]/2] - 8 Cos[Subscript[k, x]/2]^3 - Sqrt[
81 Cos[Subscript[k, x]/2]^2 -
2 \[Pi]^2 Cos[Subscript[k, x]/2]^2]) Sec[Subscript[k, x]/
2]^2])/Sqrt[3]), {Subscript[k, x], -10, 10}]


with a plot that doesn't look right at all:

In the end I expect something that is comparable to this:

How can I plot sections through plane (cuts through levels), so that I get curves in 2D-space?

• What is B2 supposed to be? – J. M. is in limbo Mar 27 '18 at 11:17
• Sorry all ´B2´ is ´B´. I correct that. – Leviathan Mar 27 '18 at 11:18
• You can still use ContourPlot[] for plotting isolated slices: ContourPlot[B[(Cos[kx] + 2 Cos[kx/2] Cos[Sqrt[3] kz/2] + 6)/2] == π/2, {kx, -6 π/3, 6 π/3}, {kz, -6 π/3, 6 π/3}]. – J. M. is in limbo Mar 27 '18 at 11:21
• Ah so simple, so true. :D – Leviathan Mar 27 '18 at 11:22

ContourPlot[