Pt ⇔4cos22x−(2−√3)cos2x−2−√3=sin2x⇔4(cos22x−1)−(2−√3)(cos2x+1)=sin2x
⇔(cos2x+1)(4cos2x−2−√3)−2sinxcosx=0
⇔cos2x(4cos2x−2−√3)−sinxcosx=0
⇔cosx(4cosx.cos2x−(2+√3)cosx−sinx)=0
⇔cosx=0∨4cosxcos2x−(2+√3)cosx−sinx=0(1)
(1)⇔4cosx(2cos2x−1)−(2+√3)cosx−sinx=0
⇔8cos3x−(6+√3)cosx−sinx=0
Chia cho cos3x ta được:
8−(6+√3)(1+tan2x)−tanx(1+tan2x)=0
Từ giải nốt