So I made a table in which I would like to have a curly bracket, containing three formulas in one cell. I managed to make one that look right in the pdf, tho I still get an error message.

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So this is what I want, and also what I get when I export to pfd. But there are a few errors that come up which I don't know how to fix.

        \begin{tabular}{l l}
            Polaire vergelijking & Cartesiaans equivalent\\
            $r \cos\theta = 2$ & $x=2$\\
            $r^2\cos \theta \sin \theta = 4$ & $xy=4$\\
            $r^2\cos^2\theta-r^2\sin^2\theta=1$ & $x^2-y^2=1$\\
            $r=1+2r\cos\theta$ & \left\{ \begin{tabular} {@{}l@{}} $y^2-3x^2-4x-1=0$\\  $\sqrt{x^2+y^2}=1+2x$\\ $x^2+y^2=(1+2x)^2$ \end{tabular} \right.\\
  • 1
    \left\{ needs math mode so you need $ as in your other math mode cells – David Carlisle Dec 27 '20 at 18:25
  • Do you mean around the entire \left\{ ? Because when I do that the bracket becomes a smaller one. So I'm not entirely sure what you mean. – robbestephan Dec 27 '20 at 18:45
  • @robbestephan I think @David Carlisle $\left\{ .......... \right.$ meant this. – Imran Dec 27 '20 at 19:02
  • Ok, makes sense – robbestephan Dec 27 '20 at 19:20

You may liked:

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Instead nested tabular is used cases environment defined in the amsmath package. For more vertical space is added makecell package to preamble and in code for table used its macro \makegapedcells. Instead tabular is used array and it is enclosed in math environment. By this you not need to write $ before and after each equations in table:

\usepackage{amsmath}  % <---
\usepackage{makecell} % <---

\setcellgapes{3pt} % <---
\makegapedcells    % <---
    \begin{array}{l l}
    $Polaire vergelijking$ & $Cartesiaans equivalent$   \\
r \cos\theta = 2                    & x=2           \\
r^2\cos \theta \sin \theta = 4      & xy=4          \\
r^2\cos^2\theta-r^2\sin^2\theta=1   & x^2-y^2=1     \\
r=1+2r\cos\theta                    & \begin{cases} 
                            y^2-3x^2-4x-1=0         \\  
                            \sqrt{x^2+y^2}=1+2x     \\ 
                                      \end{cases}   \\
  • Thank you, this helped a lot. – robbestephan Dec 27 '20 at 18:44
  • If answer solve your problem, you can accept it (by clicking on check mark at top left side of answer). – Zarko Dec 27 '20 at 18:48

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