lấy máy tính fx-570ES plus ra mà bấm

Thế thỳ cần j phải đăng lên dây hỏi

\(\int\limits\frac{\sin2x}{3+4\sin x-\left(1-2\sin^2x^{ }\right)}dx\)=\(\int\limits^{\Pi}_2\frac{2\sin x\cos x}{2\left(\sin^2x+2\sin x+1\right)}dx\)=\(\int\limits^{\frac{\Pi}{2}}_0\frac{\sin x\cos x}{\left(\sin x+1\right)^2}dx\)  đặt \(\sin x+1=t\Rightarrow\cos xdx=dt\)  đổi cận x và t 

→\(\int\limits^2_1\frac{\left(t^2-1\right)}{t^2}dt=\int\limits^2_1\left(1-\frac{1}{t^2}\right)dt\)=(t+\(\frac{1}{t}\)) thế cận vào tính là ok 

a/

\(\sqrt{3}sin4x-cos4x=2cosx\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=cosx\)

\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{6}=\frac{\pi}{2}-x+k2\pi\\4x-\frac{\pi}{6}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)

b/

\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)

\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=sin\left(2x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)

\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

\(2\sqrt{3}sinx.cosx+1-2sin^2x=1-4sinx\)

\(\Leftrightarrow\sqrt{3}sinx.cosx-sin^2x+2sinx=0\)

\(\Leftrightarrow sinx\left(\sqrt{3}cosx-sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sin\left(x-\frac{\pi}{3}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Vậy nghiệm dương nhỏ nhất \(x=\frac{5\pi}{6}\Rightarrow a+b=11\)

\(\left|cosx\right|-\left|sinx\right|-\left(\left|cosx\right|-\left|sinx\right|\right)\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}=0\)

\(\Leftrightarrow\left(\left|cosx\right|-\left|sinx\right|\right)\left(1-\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|cosx\right|=\left|sinx\right|\Leftrightarrow cos2x=0\left(1\right)\\\left(\left|cosx\right|+\left|sinx\right|\right)\sqrt{1+sin2x}=1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

\(\left(2\right)\Leftrightarrow\left|cosx\right|+\left|sinx\right|=\dfrac{1}{\sqrt{1+sin2x}}\) (với \(sin2x\ne-1\))

\(\Leftrightarrow1+2\left|sinx.cosx\right|=\dfrac{1}{1+sin2x}\)

\(\Leftrightarrow1+\left|sin2x\right|=\dfrac{1}{1+sin2x}\)

TH1: \(-1< sin2x< 0\Rightarrow1-sin2x=\dfrac{1}{1+sin2x}\)

\(\Leftrightarrow1-sin^22x=1\Rightarrow sin2x=0\) (loại)

TH2: \(0\le sin2x\le1\Rightarrow1+sin2x=\dfrac{1}{1+sin2x}\)

\(\Leftrightarrow1+sin2x=1\Leftrightarrow sin2x=0\Rightarrow x=\dfrac{k\pi}{2}\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)

Bạn tự tìm số giá trị nhé

@Nguyễn Việt Lâm giúp em với