Parenthesis disappears when I use align environment

Question

I have an long equation that I must break. However, when I do so, a set of parenthesis disappears. Furthermore, the formation of the equation is odd. What should I do? As you will see in the image, the first equation seems ok, but it's the second one. The red marking shows the missing parenthesis. Thank you.

\documentclass[11pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}

\begin{document}

\begin{alignat}{2}
\begin{aligned}
&\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \Dot{x}}\right)-\frac{\partial \mathcal{L}}{\partial x}=F &&\Rightarrow \left(2m_0+2m_1+2m_2\right)\Ddot{x}+\left(L_1m_1\cos\theta_2+2L_1m_2\cos\theta_1\right)\Ddot{\theta}_1\\
&\:&&+\left(L_1m_1\sin\theta_1+2L_1m_2\sin\theta_1\right)\Dot{\theta}_1^2+\left(L_2m_2\sin\theta_2\right)\Dot{\theta}_2^2=2F\\
%%%%%%%%%%%%SECOND EQUATION%%%%%%%%%%%%
&\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial\Dot{\theta}_1}\right)-\frac{\partial\mathcal{L}}{\partial\theta_1}=0&&\Rightarrow\left(12\,m_{1}\,\cos\theta_1+24\,m_{2}\,\cos\theta_1\right)\Ddot{x}+\left(2\,m_{1}+15\,L_{1}\,m_{1}\\
 &\:&&+24\,L_{1}\,m_{2}-9\,L_{1}\,m_{1}\,\cos\left(2\,\theta_{1}\right)\right)\Ddot{\theta}_1+\left(12\,L_{2}\,m_{2}\,\cos\left(\theta_{1}-\theta_{2}\right)\right)\Ddot{\theta}_2\\
 &\:&&+\left(9\,L_{1}\,m_{1}\,\sin\left(2\,\theta _{1}\right)\right)\Dot{\theta}_1^2+\left(12\,L_{2}\,m_{2}\,\sin\left(\theta _{1}-\theta _{2}\right)\right)\Dot{\theta}_2^2\\
 &\:&&-12\,g\,m_{1}\,\sin\theta _{1}=0
\end{aligned}
\end{alignat}

\end{document} 

Answer 1

As @UlrikeFischer has already noted in a comment, your code contains a syntax error, as \left and \right mustn't span line breaks.

The simplest fix is to get rid of almost all \left and \right sizing modifiers, especially since they don't actually do anything (other than to create a syntax error.)

\documentclass[11pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsmath,amssymb}
\usepackage{geometry}

\begin{document}

\begin{align}
\smash[b]{\frac{d}{dt}
 \left(\frac{\partial \mathcal{L}}{\partial \Dot{x}}\right)
 -\frac{\partial \mathcal{L}}{\partial x}=F }\Rightarrow{}
& (2m_0+2m_1+2m_2)\Ddot{x}
 +(L_1m_1\cos\theta_2+2L_1m_2\cos\theta_1)\Ddot{\theta}_1\notag\\
&\quad+
  (L_1m_1\sin\theta_1+2L_1m_2\sin\theta_1)\Dot{\theta}_1^2
+(L_2m_2\sin\theta_2)\Dot{\theta}_2^2\notag\\
&= 2F\\
%%
\smash[b]{\frac{d}{dt}
 \left(\frac{\partial \mathcal{L}}{\partial\Dot{\theta}_1}\right)
 -\frac{\partial\mathcal{L}}{\partial\theta_1}=0}\Rightarrow{}
&(12m_{1}\cos\theta_1+24m_{2}\cos\theta_1)\Ddot{x}
 +\bigl[2m_{1}+15L_{1}m_{1}+24L_{1}m_{2}\notag\\
&\quad-\phantom{(}9L_{1}m_{1}
 \cos(2\theta_{1})\bigr]\Ddot{\theta}_1
+(12L_{2}m_{2}\cos(\theta_{1}-\theta_{2}))\Ddot{\theta}_2\notag\\
&\quad+(9L_{1}m_{1}\sin(2\theta _{1}))\Dot{\theta}_1^2
      +(12L_{2}m_{2}\sin(\theta _{1}-\theta _{2}))\Dot{\theta}_2^2\notag\\
&\quad-12gm_{1}\sin\theta _{1}\notag\\
&=0
\end{align}

\end{document} 

Answer 2

I propose this variant layout, to avoid overfull boxes. It requires the nccmath package (to be loaded before mathtools). I removed the unnecessary \left...\right pairs, and replaced the others with the bigl ... \bigr pair. Last, I used the diffcoeff package to simplify the typing of derivatives in Leibniz notation:

\documentclass[11pt,a4paper]{article}
\usepackage[english]{babel}
\usepackage{nccmath}
\usepackage{mathtools}
\usepackage{amssymb}
\usepackage{diffcoeff}

\begin{document}

\begin{fleqn}
\begin{align}
 & \begin{alignedat}{2}%
  & \diff*{\biggl(\diffp{\mathcal{L}}{\Dot{x}}\biggr)}{t} & & -\diffp{\mathcal{L}}{x} = F\Rightarrow \\
 & & & \phantom{{}+{}}(2m_0+2m_1+2m_2)\Ddot{x}+(L_1m_1\cos\theta_2+2L_1m_2\cos\theta_1)\Ddot{\theta}_1\\
 & & & +(L_1m_1\sin\theta_1+2L_1m_2\sin\theta_1)\Dot{\theta}_1^2+(L_2m_2\sin\theta_2)\Dot{\theta}_2^2=2F
\end{alignedat}\\[2ex]
%%%%%%%%%%%%%SECOND EQUATION%%%%%%%%%%%%
 & \begin{alignedat}{2}
 & \diff*{\biggl(\diffp{\mathcal{L}}{\Dot{\theta_1}}\biggr)}{t} & & -\diffp{\mathcal{L}}{\theta_1}=0\Rightarrow \\%
 & & &\phantom{{}+{}}(12\,m_{1}\,\cos\theta_1+ 24\,m_{2}\,\cos\theta_1)\Ddot{x}\\ %
 & & & + \bigl(2\,m_{1}+15\,L_{1}\,m_{1}+24\,L_{1}\,m_{2}-9\,L_{1}\,m_{1}\,\cos 2\,\theta_{1}\bigr)\Ddot{\theta}_1\\
 & & & +\bigl(12\,L_{2}\,m_{2}\,\cos(\theta_{1}-\theta_{2})\bigr)\Ddot{\theta}_2 +(9\,L_{1}\,m_{1}\,\sin 2\,\theta _{1})\Dot{\theta}_1^2\\
 & & & +\bigl(12\,L_{2}\,m_{2}\,\sin(\theta _{1}-\theta _{2})\bigr)\Dot{\theta}_2^2-12\,g\,m_{1}\,\sin\theta _{1}=0
 \end{alignedat}
\end{align}
 \end{fleqn}

\end{document}