Ta có: \(\sin xcosx+cosx=-2\cdot\sin^2x-\sin x+1\)
=>cosx(sin x+1)=\(-2\cdot\sin^2x-2\cdot\sin x+\sin x+1\)
=>(sin x+1)(cosx+2*sin x-1)=0
TH1: sin x+1=0
=>sin x=-1
=>\(x=-\frac{\pi}{2}+k2\pi\)
TH2: 2sin x+cosx-1=0
=>\(\frac{2}{\sqrt5}\cdot\sin x+\frac{1}{\sqrt5}\cdot cosx=\frac{1}{\sqrt5}\)
=>sin(x+α)=sinα
=>\(\left[\begin{array}{l}x+\alpha=\alpha+k2\pi\\ x+\alpha=\pi-a+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=k2\pi\\ x=\pi-2\alpha+k2\pi\end{array}\right.\)
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