a.
\(\Leftrightarrow2sinx.cosx+4cos^2x-2=1+sinx-4cosx\)
\(\Leftrightarrow2sinx.cosx-sinx+\left(4cos^2x+4cosx-3\right)=0\)
\(\Leftrightarrow sinx\left(2cosx-1\right)+\left(2cosx-1\right)\left(2cosx+3\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+2cosx+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+2cosx+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\Rightarrow x=...\\sinx+2cosx=-3\left(1\right)\end{matrix}\right.\)
Xét (1), ta có \(1^2+2^2< \left(-3\right)^2\) nên (1) vô nghiệm (theo đk có nghiệm của pt lượng giác bậc nhất)
b.
\(\Leftrightarrow4sinx.cosx-1+2sin^2x=7sinx+2cosx-4\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(2sin^2x-7sinx+3\right)=0\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cosx+sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\Rightarrow x\\2cosx+sinx=3\left(1\right)\end{matrix}\right.\)
Xét (1)
Do \(2^2+1^2< 3^2\) nên (1) vô nghiệm
c.
\(cos^2x-sin^2x+\left(1+2cosx\right)\left(sinx-cosx\right)=0\)
\(\Leftrightarrow-\left(sinx+cosx\right)\left(sinx-cosx\right)+\left(1+2cosx\right)\left(sinx-cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(1+2cosx-sinx-cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(cosx-sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\sinx-cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\\\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow\left(sinx-cosx+1\right)\left(2sinx-cosx\right)-2sinx.cosx=0\)
\(\Leftrightarrow\left(sinx-cosx+1\right)\left(2sinx-cosx\right)-2sinx.cosx+1-1=0\)
\(\Leftrightarrow\left(sinx-cosx+1\right)\left(2sinx-cosx\right)+sin^2x+cos^2x-2sinx.cosx-1=0\)
\(\Leftrightarrow\left(sinx-cosx+1\right)\left(2sinx-cosx\right)+\left(sinx-cosx\right)^2-1=0\)
\(\Leftrightarrow\left(sinx-cosx+1\right)\left(2sinx-cosx\right)+\left(sinx-cosx+1\right)\left(sinx-cosx-1\right)=0\)
\(\Leftrightarrow\left(sinx-cosx+1\right)\left(3sinx-2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx+1=0\\3sinx-2cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-1\\\frac{3}{\sqrt{13}}sinx-\frac{2}{\sqrt{13}}cosx=\frac{1}{\sqrt{13}}\end{matrix}\right.\)
Đặt \(cosa=\frac{3}{\sqrt{13}}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\sin\left(x-a\right)=\frac{1}{\sqrt{13}}\end{matrix}\right.\)
\(\Leftrightarrow...\)
