(The question has changed since the answer was posted.)
Hi,
is this a Mathematica question or a general Mathematics question?
The equation you give is clearly not in Mathematica format and we need to make several assumptions here to guess what you are looking for. Let’s suppose that you are looking for a curve given in polar coordinates, where r is the radius and \Theta is the angle. We could then plot the curve
PolarPlot[Sqrt[Cos[theta]^2 + 1], {theta, 0, 2 Pi}, AspectRatio -> 1]
It does look somewhat nicer if you square the \Theta only and not the Cos:
PolarPlot[Sqrt[Cos[theta^2] + 1], {theta, 0, 2 Pi}, AspectRatio -> 1]
We can also rotate for more than 2 Pi and add a little color:
PolarPlot[Sqrt[Cos[theta^2] + 1], {theta, 0, 16 Pi}, AspectRatio -> 1,PlotPoints -> 500, ColorFunction -> Function[{x, y, t, r}, Hue[r]]]
Well, there are lot’s of things you can do with Mathematica. It is worthwhile to have a look at Mathematica. You can, of course, also just type your equation into Wolfram Alpha like so:
http://www.wolframalpha.com/input/?i=r%3D%3DSqrt%5BCos%5Btheta%5D%5E2%2B1%5D
That gives you a whole lot of information about that implicit function. Here are the first couple of lines from the output:
You even get sliders etc, to better study the equation.
Cheers,
M.
Here the implicit equation, it is a case of Booth oval.
Thank you for your cooperation.
If you want the solution in cartesian coordinates, which now appears what you want, you can use:
TransformedField["Polar" -> "Cartesian", r == Sqrt[Cos[\[CapitalTheta]]^2 + 1], {r, \[Theta]} -> {x, y}]
(*Sqrt[x^2 + y^2] == Sqrt[1 + Cos[\[CapitalTheta]]^2]*)
That still contains the Theta so you can use
TransformedField["Polar" -> "Cartesian", r == Sqrt[Cos[\[CapitalTheta]]^2 + 1], {r, \[Theta]} -> {x, y}] /. \[CapitalTheta] ->
CoordinateTransform["Cartesian" -> "Polar", {x, y}][[2]]//FullSimplify
(*x^2 + y^2 == 1 + x^2/(x^2 + y^2)*)
To convert that to the equation you desire, you can multiply both sides by (x^2+y^2):
Map[#*(x^2 + y^2) &, %] // FullSimplify
(* (x^2 + y^2)^2 == 2 x^2 + y^2*)
Cheers,
M.