\begin{frame}{Phương trình $\sin x=m$}\pause
\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
\[\boxed{\sin x=m \Leftrightarrow \sin x=\sin \alpha \pause \Leftrightarrow \left[\begin{aligned}&x=\alpha+k2\pi, k\in\mathbb{Z} \pause \\&x=\pi-\alpha+k2\pi,k\in\mathbb{Z}.\end{aligned}\right.}\]
\begin{tikzpicture}[scale=.8,font=\footnotesize,line cap=round,line join=round,>=stealth]
\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]{$\cos$};

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

1 Answer 1


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.

enter image description here

% removing [handout] to use \pause
\begin{frame}[t]{$\sin x=m$}\pause
\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
\draw[->] (-1.5,0)--(1.5,0) node[above]{$\cos$};
\draw[->] (0,-1.5)--(0,1.5) node[right]{$\sin$};
(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);
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 
&x={\color{red}\alpha}+k2\pi, \quad k\in\mathbb{Z} \\ \pause
&x={\color{red}\pi-\alpha}+k2\pi,\quad k\in\mathbb{Z}.

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