multiple column align equation environments

Question

Can I get this three column arrangement in the equation environment without using three align environments? The problem I am having is the amsmath error "erroneous nesting of equations" occurs and I want to avoid that error/warning. I also don't want to have to write out all the first lines together, then all the second lines etc. etc. because that would get confusing when editing the equations later on. The following MWC is complex on purpose. It illustrates how confusing it would get if I had to write all the lines as described above. thanks.

\documentclass[]{article}
\usepackage{amsmath}
\begin{document}

\begin{equation*}
\begin{align}
P_{S}   &= V_{S} \times I_{1} \times cos(\phi) \\
        &=  \times  \times cos(\phi) \\
\therefore     cos(\phi)_{1A}  &= \frac{}{ \times } \\
            &=  \\
I_{S}           &= I_{1} \sqrt{1+THD^{2}} \\
\therefore \ 
THD     &= \frac{I_{S}}{I_{1}} - 1\\    
        &= \frac{}{} - 1 \\
        &=  \\
df      &= cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} \\               
        &=  \times \frac{1}{\sqrt{1+^{2}}} \\
        &=      
\end{align}
\begin{align}
P_{S}   &= V_{S} \times I_{1} \times cos(\phi) \\
    &=  \times  \times cos(\phi) \\
\therefore     cos(\phi)_{1A}  &= \frac{}{ \times } \\
            &=  \\
I_{S}           &= I_{1} \sqrt{1+THD^{2}} \\
\therefore \ 
THD     &= \frac{I_{S}}{I_{1}} - 1\\    
        &= \frac{}{} - 1 \\
        &=  \\
df      &= cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} \\               
        &=  \times \frac{1}{\sqrt{1+^{2}}} \\
        &= \\
\end{align}
\begin{align}
P_{S}   &= V_{S} \times I_{1} \times cos(\phi) \\
    &=  \times  \times cos(\phi) \\
\therefore     cos(\phi)_{1A}  &= \frac{}{ \times } \\
            &=  \\
I_{S}           &= I_{1} \sqrt{1+THD^{2}} \\
\therefore \ 
THD     &= \frac{I_{S}}{I_{1}} - 1\\    
        &= \frac{}{} - 1 \\
        &=  \\
df      &= cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} \\               
        &=  \times \frac{1}{\sqrt{1+^{2}}} \\
        &= \\
\end{align}
\begin{equation*}

\end{document}

Answer 1

You cannot nest an align environment inside an another display-math environment, such as equation*. Use the lower-level aligned environment, instead.

Also, \therefore is defined in amssymb; don't forget to load that package.

Note: I've chosen landscape orientation because your equations don't fit on an A4 page in portrait orientation with default margins.

\documentclass{article}

\usepackage[landscape]{geometry}
\usepackage{amsmath,amssymb}

\begin{document}
\begin{equation*}
\begin{aligned}
P_{S}   &= V_{S} \times I_{1} \times cos(\phi) \\
        &=  \times  \times cos(\phi) \\
\therefore     cos(\phi)_{1A}  &= \frac{}{ \times } \\
            &=  \\
I_{S}           &= I_{1} \sqrt{1+THD^{2}} \\
\therefore \ 
THD     &= \frac{I_{S}}{I_{1}} - 1\\    
        &= \frac{}{} - 1 \\
        &=  \\
df      &= cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} \\               
        &=  \times \frac{1}{\sqrt{1+^{2}}} \\
        &=      
\end{aligned}
\begin{aligned}
P_{S}   &= V_{S} \times I_{1} \times cos(\phi) \\
    &=  \times  \times cos(\phi) \\
\therefore     cos(\phi)_{1A}  &= \frac{}{ \times } \\
            &=  \\
I_{S}           &= I_{1} \sqrt{1+THD^{2}} \\
\therefore \ 
THD     &= \frac{I_{S}}{I_{1}} - 1\\    
        &= \frac{}{} - 1 \\
        &=  \\
df      &= cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} \\               
        &=  \times \frac{1}{\sqrt{1+^{2}}} \\
        &= \\
\end{aligned}
\begin{aligned}
P_{S}   &= V_{S} \times I_{1} \times cos(\phi) \\
    &=  \times  \times cos(\phi) \\
\therefore     cos(\phi)_{1A}  &= \frac{}{ \times } \\
            &=  \\
I_{S}           &= I_{1} \sqrt{1+THD^{2}} \\
\therefore \ 
THD     &= \frac{I_{S}}{I_{1}} - 1\\    
        &= \frac{}{} - 1 \\
        &=  \\
df      &= cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} \\               
        &=  \times \frac{1}{\sqrt{1+^{2}}} \\
        &= \\
\end{aligned}
\end{equation*}
\end{document}

Answer 2

I think the other answer is what you want, but just for reference, here's how you can do it in one big align environment- with a bit of code tidying it doesn't look too bad

% arara: pdflatex
% !arara: indent: {overwrite: on}
\documentclass{article}
\usepackage[landscape]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}

\begin{align}
    P_{S}                        & = V_{S} \times I_{1} \times cos(\phi)         & P_{S}                     & = V_{S} \times I_{1} \times cos(\phi)         & P_{S}                       & = V_{S} \times I_{1} \times cos(\phi)         \\
                                 & =  \times  \times cos(\phi)                   &                           & =  \times  \times cos(\phi)                   &                             & =  \times  \times cos(\phi)                   \\
    \therefore \  cos(\phi)_{1A} & = \frac{}{ \times }                           & \therefore cos(\phi)_{1A} & = \frac{}{ \times }                           & \therefore \ cos(\phi)_{1A} & = \frac{}{ \times }                           \\
                                 & =                                             &                           & =                                             &                             & =                                             \\
    I_{S}                        & = I_{1} \sqrt{1+THD^{2}}                      & I_{S}                     & = I_{1} \sqrt{1+THD^{2}}                      & I_{S}                       & = I_{1} \sqrt{1+THD^{2}}                      \\
    \therefore \  THD            & = \frac{I_{S}}{I_{1}} - 1                     & \therefore  THD           & = \frac{I_{S}}{I_{1}} - 1                     & \therefore \ THD            & = \frac{I_{S}}{I_{1}} - 1                     \\    
                                 & = \frac{}{} - 1                               &                           & = \frac{}{} - 1                               &                             & = \frac{}{} - 1                               \\
                                 & =                                             &                           & =                                             &                             & =                                             \\
    df                           & = cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} & df                        & = cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} & df                          & = cos(\phi) \times \frac{1}{\sqrt{1+THD^{2}}} \\               
                                 & =  \times \frac{1}{\sqrt{1+^{2}}}             &                           & =  \times \frac{1}{\sqrt{1+^{2}}}             &                             & =  \times \frac{1}{\sqrt{1+^{2}}}             \\
                                 & =                                             &                           & =                                             &                             & =                                             \\
\end{align}
\end{document}