3tan2x−4√3|tanx|+4secx+11=0
⇒2tan2x+tan2x+1−4√3|tanx|+4secx+10=0
⇒2(tan2x−2√3|tanx|+3)+(sec2x+4secx+4)=0
⇒2(|tanx|−√3)2+(secx+2)2=0
This is possible only when
⇒|tanx|=√3 and secx+2=0
⇒tanx=±√3 and secx=−2
⇒tanx=±√3 and cosx=−12⇒x=π3,2π3,4π3,5π3 and x=2π3,4π3
∴x=2π3,4π3
Hence, the number of solutions is 2.