2√2sin(x−π12)cosx=1
⇔2√2sin(π4+x−π3)cosx=1
⇔2[sin(x−π3)+cos(x−π3)]cosx=1
⇔2[sinxcosπ3−cosxsinxπ3+cosxcosπ3+sinxsinπ3]cosx=1
⇔(sinx−√3cosx+cosx+√3sinx)cosx=1
⇔[(1+√3)sinx+(1−√3)cosx]cosx=1
⇔(1+√3)sinxcosx+(1−√3)cos2x=sin2x+cos2x
⇔(1+√3)tanx−√3=tan2x
⇔tan2x−(1+√3)tanx+√3=0
⇔tanx=1 hoặc tanx=√3