ĐKXĐ: \(2\cdot cosx+\sqrt3<>0\)
=>\(cosx<>-\frac{\sqrt3}{2}\)
=>\(x<>\pm\frac56\pi+k2\pi\)
Ta có: \(\frac{\sin2x-2\cdot cos^2x-5\cdot\sin x-cosx+4}{2\cdot cosx+\sqrt3}=0\)
=>\(\sin2x-2\cdot cos^2x-5\cdot\sin x-cosx+4=0\)
=>\(\left(\sin2x-cosx\right)+\left(-2\cdot cos^2x+4\right)-5\cdot\sin x=0\)
=>\(\left(2\cdot\sin x\cdot cosx-cosx\right)+2\cdot\left(2-cos^2x\right)-5\cdot\sin x=0\)
=>\(cosx\cdot\left(2\sin x-1\right)+2\cdot\left(1+\sin^2x\right)-5\cdot\sin x=0\)
=>\(cosx\cdot\left(2\cdot\sin x-1\right)+2\cdot\sin^2x-5\cdot\sin x+2=0\)
=>\(cosx\cdot\left(2\cdot\sin x-1\right)+\left(2\cdot\sin x-1\right)\left(\sin x-2\right)=0\)
=>\(\left(2\cdot\sin x-1\right)\left(cosx+\sin x-2\right)=0\)
TH1: 2sin x-1=0
=>2 sin x=1
=>\(\sin x=\frac12\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\left(nhận\right)\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\left(loại\right)\end{array}\right.\)
TH2: \(cosx+\sin x-2=0\)
=>\(\sin x+cosx=2\)
=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=2\)
=>\(\sin\left(x+\frac{\pi}{4}\right)=\sqrt2>1\)
=>LOại
Vậy: \(x=\frac{\pi}{6}+k2\pi\)
