Đặt \(t=\sin2x+cos2x\)

=>\(t^2=\left(\sin2x+cos2x\right)^2=1+2\cdot\sin2x\cdot cos2x\)

=>\(2\cdot\sin2x\cdot cos2x=t^2-1\)

=>\(\sin2x\cdot cos2x=\frac{t^2-1}{2}\)

Ta có: \(t=\sin2x+cos2x\)

=>\(t=\sqrt2\cdot\sin\left(2x+\frac{\pi}{4}\right)\)

Ta có: \(-1<=\sin\left(2x+\frac{\pi}{4}\right)\le1\)

=>\(-\sqrt2\le\sqrt2\cdot\sin\left(2x+\frac{\pi}{4}\right)\le\sqrt2\)

=>\(-\sqrt2\le t\le\sqrt2\)

Ta có: \(\left(\sin2x+cos2x\right)\left(1-\sin2x\cdot cos2x\right)+\sin2x\cdot cos2x=1\)

=>\(t\cdot\left(1-\frac{t^2-1}{2}\right)+\frac{t^2-1}{2}=1\)

=>\(t\cdot\frac{2-t^2+1}{2}+\frac{t^2-1}{2}=1\)

=>\(\frac{-t^3+3t+t^2-1}{2}=1\)

=>\(-t^3+t^2+3t-1=2\)

=>\(t^3-t^2-3t+1=-2\)

=>\(t^3-t^2-3t+3=0\)

=>\(\left(t-1\right)\left(t^2-3\right)=0\)

=>\(\left[\begin{array}{l}t-1=0\\ t^2-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}t=1\\ t^2=3\end{array}\right.\Rightarrow\left[\begin{array}{l}t=1\left(nhận\right)\\ t=\sqrt3\left(loại\right)\\ t=-\sqrt3\left(loại\right)\end{array}\right.\)

=>t=1

=>\(\sqrt2\cdot\sin\left(2x+\frac{\pi}{4}\right)=1\)

=>\(\sin\left(2x+\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\ 2x+\frac{\pi}{4}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=k2\pi\\ 2x=\frac12\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=k\pi\\ x=\frac{\pi}{4}+k\pi\end{array}\right.\)

Chọn D

∫ sin 2 x d x cos 2 x - 1 = ∫ 2 sin x cos x 1 - 2 sin 2 x + 1 d x   = ∫ 2 sin x . cos x 2 cos 2 x d x = ∫ sin x cos x d x   = - ∫ d ( cos x ) cos x = - ln cos x + C

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{cosx}{sinx}-1=\dfrac{cos^2x-sin^2x}{1+\dfrac{sinx}{cosx}}+sin^2x-sinx.cosx\)

\(\Leftrightarrow\dfrac{cosx-sinx}{sinx}=cosx\left(cosx-sinx\right)-sinx\left(cosx-sinx\right)\)

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

\(\Leftrightarrow\left(cosx-sinx\right)\left(1-sinx.cosx+sin^2x\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-sin2x-cos2x\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\right)=0\)

\(cos2x+cosx+1=sin2x+sinx\)

\(\Leftrightarrow cos^2x-sin^2x+cosx+cos^2x+sin^2x=2sinx.cosx+sinx\)

\(\Leftrightarrow2cos^2x+cosx=2sinx.cosx+sinx\)

\(\Leftrightarrow cosx\left(2cosx+1\right)=sinx\left(2cosx+1\right)\)

\(\Leftrightarrow\left(2cosx+1\right)\left(sinx-cosx\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\tanx=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{4}+k\pi\\\end{matrix}\right.\)

Chọn A

\(sin2x-cos2x+3sinx-cosx-1=0\)

\(\Leftrightarrow2sinxcosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)

\(\Leftrightarrow2sinxcosx-1+2sin^2x+3sinx-cosx-1=0\)

\(\Leftrightarrow2sin^2x+3sinx-2+cosx\left(2sinx-1\right)=0\)

\(\Leftrightarrow2\left(sinx-\dfrac{1}{2}\right)\left(sinx+2\right)+cosx\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+2\right)+cosx\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+2+cosx\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx+cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=sin\dfrac{\pi}{6}\\\sqrt[]{2}\left(sinx.\dfrac{1}{\sqrt[]{2}}+cosx.\dfrac{1}{\sqrt[]{2}}\right)=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\\sqrt[]{2}sin\left(x+\dfrac{\pi}{4}\right)=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt[]{2}\left(vô.lý\right)\end{matrix}\right.\)

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

sin 2   x   -   cos 2   x   =   cos 4 x   ⇔   - cos 2 x   =   cos 4 x   ⇔   2 cos 3 x . cos x   =   0

\(\left(2cos2x-1\right)\left(sin2x+cos2x\right)=1\)

\(\Leftrightarrow2sin2x.cos2x+2cos^22x-sin2x-cos2x-1=0\)

\(\Leftrightarrow sin4x+cos4x-sin2x-cos2x=0\)

\(\Leftrightarrow2cos3x.sinx-2sin3x.sinx=0\)

\(\Leftrightarrow2sinx\left(cos3x-sin3x\right)=0\)

\(\Leftrightarrow2\sqrt{2}sinx.cos\left(3x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos\left(3x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\3x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\end{matrix}\right.\)