1−2cos2x−√3sinx+cosx=0⇔1−2(2cos2x−1)+cosx−√3sinx=0
⇔−4cos2x+cosx+3−√3sinx=0
⇔(1−cosx)(4cosx+3)−√3sinx=0
⇔2sin2x2(4cosx+3)−2√3sinx2cosx2=0⇔[2sinx2=0(1)sinx2(4cosx+3)−√3cosx2=0(2)
(1)⇔x2=kπ⇔x=k2π(k∈Z)
(2)⇔4cosxsinx2+3sinx2−√3cosx2=0
⇔2sin3x2−2sinx2+3sinx2−√3cosx2=0
⇔sin3x2=√32cosx2−12sinx2
⇔sin3x2=sin(π3−x2)
⇔[3x2=π3−x2+k2π3x2=2π3+x2+k2π
⇔[x=π6+kπx=2π3+k2π (k∈Z)