$cos^{2}x+$ $3sin^{2}x -sin2x+2cosx+4sinx-7=0$
⇔ $cos^{2}x+$ $3(1-cos^{2}x)-2sinxcosx+2cosx+4sinx-7=0$
⇔$-2cos^{2}x-2sinxcosx+2cosx+4sinx-4=0$
⇔$-cos^{2}x-sinxcosx+cosx+2sinx-2=0$
⇔$-(1-sin^{2}x)-sinxcosx+cosx+2sinx-2=0$
⇔$(sinx+1)(sinx-1)-cosx(sinx-1)+2(sinx-1)=0$
⇔$(sinx-1)[(sinx+1)-cosx+2]=0$
⇔\(\left[ \begin{array}{l}sinx-1=0\\sinx+1-cosx+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}sinx=-1\\sinx-cosx=-3(VN)\end{array} \right.\)
⇔$x=$$\frac{-\pi}{2}+k2\pi$ $(k∈Z)$
