\documentclass{beamer} \usepackage[utf8x]{vietnam} \usepackage{ucs} \usepackage{amsmath} \usepackage{amsfonts}\usetheme{Warsaw} \usepackage{amssymb,enumerate}\setbeamertemplate{background canvas}[vertical shading][bottom=yellow!15,top=red!10] \usepackage{graphicx} \hypersetup{pdfpagemode=FullScreen, bookmarks=false,unicode,bookmarksopen=false,unicode} \usepackage{multicol} \title{PHƯƠNG TRÌNH LƯỢNG GIÁC CƠ BẢN} \author{GV: Nguyễn Văn Minh Hiếu} \date{\today} \begin{document} \begin{frame} \titlepage \end{frame} \begin{frame} \frametitle{ Giải các phương trình sau:} \begin{enumerate} \item<2-> $2\sin{\Big(2x-\dfrac{\pi}{3}\Big)}+\sqrt3=0 $\transblindsvertical<2,3> \item<3-> $2\cos{\Big(x+\dfrac{\pi}{4}\Big)}+\sqrt2=0 $ \item<4-> $\sin{\Big(2x-\dfrac{\pi}{4}\Big)}=\sin{3x} $ \item<5-> $\sin{\Big(2x+\dfrac{2\pi}{3}\Big)} +\sin{2x}=0$ \item<6-> $\cos{\Big(x+\dfrac{3\pi}{4}\Big)}=\cos{x} $ \item<7-> $\cos{\Big(3x-\dfrac{\pi}{4}\Big)} +\cos{2x}=0$ \item<8-> $\sin{\Big(3x+\dfrac{\pi}{3}\Big)}-\cos{x}=0 $ \item<9-> $\cos{3x}+\sin{2x}=0 $ \item<10-> $\cos^2{\Big(2x-\dfrac{\pi}{3}\Big)}-\sin^2{x}=0 $ \item<11-> $ \cos^2 x-\sin^2 x=\sin 4x$ \end{enumerate} \end{frame} \begin{frame} \frametitle{ Giải các phương trình sau:} \begin{enumerate} \item<2-> $ 2\cos^2 x-1=0$ \item<3-> $ 3-4\sin^2{2x}=0$ \item<4-> $(1-\cos x)(1+\cos x)=0 $ \item<5-> $(3-\sin x)(1-2\sin x)=0$ \item<6-> $ \sin^2{x}=\dfrac{1}{4}$ \item<7-> $\sin^2{\Big(5x+\dfrac{2\pi}{3}\Big)}=\cos^2{\Big(3x-\dfrac{\pi}{4}\Big)} $ \item<8-> $\cos{\Big(2x+\dfrac{\pi}{3}\Big)}=\cos x $ \item<9-> $\cos x=\sin \Big(3x+\dfrac{\pi}{6}\Big) $ \item<10-> $\sin \Big(2x+\dfrac{\pi}{3}\Big)= \sin \Big(\dfrac{2\pi}{3}-x\Big) $ \item<11-> $4\sin \Big(3x+\dfrac{\pi}{3}\Big)=\sqrt 6+\sqrt 2 $ \end{enumerate} \end{frame} \begin{frame} \frametitle{ Giải các phương trình sau:} \begin{enumerate} \item<2-> $\tan 3x=\tan \Big(x-\dfrac{\pi}{3}\Big) $ \item<3-> $\cot \Big(3x-\dfrac{\pi}{3}\Big)=\cot \Big(\dfrac{x}{2}+\dfrac{\pi}{6}\Big) $ \item<4-> $\tan 3x =\tan \Big(72^0-x\Big) $ \item<5-> $\tan \Big(3x+\dfrac{\pi}{6}\Big)=\cot \Big(x+\dfrac{\pi}{3}\Big) $ \item<6-> $\tan 4x.\tan x=-1 $ \item<7-> $\tan 4x=\tan \Big(\dfrac{\pi}{3}-2x\Big) $ \item<8-> $\tan \Big(3x+\dfrac{\pi}{3}\Big)=\cot \Big(\dfrac{\pi}{3}-x\Big) $ \item<9-> $\tan 3x. \tan x=1 $ \item<10-> $\tan 2x. \tan 7x =1 $ \item<11-> $ \tan \Big(x+1\Big).\cot \Big(2x+3\Big)=1$ \end{enumerate} \end{frame} \end{document}