# Image and text in beamer

\begin{frame}{Phương trình $\sin x=m$}\pause
\begin{minipage}{0.7\textwidth}
\begin{itemize}
\item Khi $|m|>1$ thì phương trình $\sin x=m$ vô nghiệm.\pause
\item Khi $|m|\leq 1$ thì phương trình $\sin x=m$ có nghiệm.\\ \pause
Nếu $\alpha$ là một nghiệm của phương trình $\sin x=m$, nghĩa là $\sin \alpha=m$, thì\pause
\boxed{\sin x=m \Leftrightarrow \sin x=\sin \alpha \pause \Leftrightarrow \left[\begin{aligned}&x=\alpha+k2\pi, k\in\mathbb{Z} \pause \\&x=\pi-\alpha+k2\pi,k\in\mathbb{Z}.\end{aligned}\right.}
\end{itemize}
\end{minipage}
\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}[scale=.8,font=\footnotesize,line cap=round,line join=round,>=stealth]
\def\r{1.5}
\draw[->] ({-\r-.5},0)--({\r+.5},0) node[below]{$x$};
\draw[->] (0,{-\r-.5})--(0,{\r+1}) node[right]{$y$};
\path (0,0) coordinate(O) (40:\r) coordinate(M) (140:\r) coordinate(M') (0,{\r+.5}) coordinate(N) (0,{\r*sin(40)}) coordinate(K);
\draw (O) circle(\r) (O)--(M)--(M')--cycle ({-\r-.5},{\r+.5})--({\r+.5},{\r+.5});
\fill (\r,0) circle(1pt) node[shift={(40:.2)}]{$1$} (0,\r) circle(1pt) node[shift={(135:.2)}]{$1$} (-\r,0) circle(1pt) node[shift={(220:.25)}]{$-1$} (0,-\r) circle(1pt) node[shift={(-45:.25)}]{$-1$} (N) circle(1pt);
\foreach \d/\g in {O/-135, M/40, M'/140, K/45} \fill (\d) circle(1pt) node[shift={(\g:.3)}]{$\d$};
\draw[->] (0:.3) arc (0:140:.3);
\draw[->] (0:.4) arc (0:40:.4);
\path (15:.6) node{$\alpha$} (0,{\r+1}) node[left]{$\sin$} ({\r+.5},0) node[right]{$\cos$};
\end{tikzpicture}
\end{minipage}
\end{frame}


I want the image on the right to be before the text on the left, how to do it?

The option [handout] ignores all \pause command. So, you just remove that option when you want to use \pause. I rearrange the text, and clean your TikZ code (the unit circle is of radius 1; to zoom out, just use [scale] for the whole TikZ picture. The simpler, the better!). The environment flushright makes the figure to the right of the minipage.

\documentclass[handout]{beamer}
% removing [handout] to use \pause
\usepackage{tikz}
\begin{document}
\begin{frame}[t]{$\sin x=m$}\pause
\begin{minipage}{0.5\textwidth}
\begin{itemize}
\item If $|m|>1$ then $\sin x=m$ has no solution.\pause\5mm] \item If |m|\leq 1 then \sin x=m has several solutions.\\ \pause \end{itemize} \end{minipage}\hfill \begin{minipage}{0.5\textwidth} \begin{flushright} \begin{tikzpicture}[scale=1.5,>=stealth] \draw[->] (-1.5,0)--(1.5,0) node[above]{\cos}; \draw[->] (0,-1.5)--(0,1.5) node[right]{\sin}; \def\goc{40} \path (0,0) coordinate (O) node[below left]{O} (\goc:1) coordinate (M) node[above right,red]{\alpha} (180-\goc:1) coordinate (M') node[above left,red]{\pi-\alpha} (1,0) node[below right]{1} (-1,0) node[below left]{-1} (0,1) node[above left]{1} (0,-1) node[below left]{-1} ; \draw[teal,thick] (O) circle(1); \draw[gray] (O)--(M)--(M')--cycle; \foreach \p in {M,M'} \fill[red] (\p) circle(1pt); \end{tikzpicture} \end{flushright} \end{minipage} Assume that \alpha is a solution of \sin x=m, that is, \sin \alpha=m. Then \pause \[\boxed{\sin x=m \Leftrightarrow \sin x=\sin \alpha \pause \Leftrightarrow \left[ \begin{aligned} &x={\color{red}\alpha}+k2\pi, \quad k\in\mathbb{Z} \\ \pause &x={\color{red}\pi-\alpha}+k2\pi,\quad k\in\mathbb{Z}. \end{aligned} \right.}
\end{frame}
\end{document}