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Đk: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m2\pi\\x\ne\dfrac{\pi}{4}+n\pi\end{matrix}\right.\left(m,n\in Z\right)\)
PT \(\Leftrightarrow1=2\sqrt{2}sinx.cosx\left(sinx-cosx\right)+2cos^2x\)
\(\Leftrightarrow\sqrt{2}.2sinx.cosx\left(sinx-cosx\right)+\left(2cos^2x-1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin2x\left(sinx-cosx\right)+\left(cosx-sinx\right)\left(cosx+sinx\right)=0\)
\(\Leftrightarrow\sqrt{2}sin2x=sinx+cosx\)
\(\Leftrightarrow\sqrt{2}sin2x=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{4}+k2\pi\\2x=\pi-x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+k\dfrac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)
a.
\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sin4x+\dfrac{\sqrt{2}}{2}cos4x=\dfrac{\sqrt{6}}{2}\)
\(\Leftrightarrow cos4x.cos\left(\dfrac{\pi}{4}\right)+sin4x.sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)
\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{4}=arccos\left(\dfrac{\sqrt{6}}{2}\right)+k2\pi\\4x-\dfrac{\pi}{4}=-arccos\left(\dfrac{\sqrt{6}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{16}-\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\end{matrix}\right.\)
b.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{3}}{6}\)
\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{3}\right)+sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\\x-\dfrac{\pi}{3}=-arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\\x=\dfrac{\pi}{3}-arrcos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\end{matrix}\right.\)
1.
\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
2.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\text{1) }3sinx-4cosx=1\\ \Leftrightarrow cos^2x+\left(\frac{4cosx+1}{3}\right)^2=1\\ \Leftrightarrow cosx=\frac{-4\pm6\sqrt{6}}{25}\\ \\ \Leftrightarrow x=arccos\left(\frac{-4\pm6\sqrt{6}}{25}\right)+k2\pi\)
\(2\text{) }\sqrt{3}sinx-cosx=1\\ \Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sinx-sin\frac{\pi}{6}\cdot cosx=\frac{1}{2}\\ \Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin\frac{\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+a2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\pi+b2\pi\end{matrix}\right.\)
\(3\text{) }\sqrt{3}cosx+sinx=-2\\ \Leftrightarrow\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=-1\\ \Leftrightarrow sin\frac{\pi}{3}\cdot cosx+cos\frac{\pi}{3}\cdot sinx=-1\\ \Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=-1=sin\frac{3\pi}{2}\\ \\ \Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{7\pi}{6}+k2\pi\)
\(4\text{) }cos4x-sin4x=1\\ \Leftrightarrow cos^24x+\left(cos4x-1\right)^2=1\\ \\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+a\pi\\4x=b2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{a\pi}{4}\\x=\frac{b\pi}{2}\end{matrix}\right.\)
\(5\text{) }\sqrt{3}cos4x+sin4x-2cos3x=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}cos4x+\frac{1}{2}sin4x=cos3x\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos4x+sin\frac{\pi}{3}\cdot sin4x=cos3x\\ \Leftrightarrow cos\left(4x-\frac{\pi}{3}\right)=cos3x\\ \Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{3}=3x+a2\pi\\4x-\frac{\pi}{3}=-3x+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\frac{\pi}{21}+\frac{b2\pi}{7}\end{matrix}\right.\\ \Leftrightarrow x=\frac{\pi}{21}+\frac{k2\pi}{7}\)
\(6\text{) }cos^2x=3sin2x+3\\ \Leftrightarrow\frac{cos2x+1}{2}=3sin2x+3\)
Giải tương tự vd 1 và 4
7) Giải tương tự vd 1 và 4
c/
\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)
\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b/
\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
1.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
Pt trở thành:
\(t^3+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow2t^3+t^2-3=0\)
\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)
Pt trở thành:
\(t^3=1+\frac{1-t^2}{2}\)
\(\Leftrightarrow2t^3+t^2-3=0\)
\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)