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)
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)