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}