Number of solutions of the equation 4cos22x+cos2x+1 +tanx(tanx−23)=0 in [0,2π] is
0
1
2
3
4cos22x+4cos2x+1+tan2x−23tanx+3=0⇒ (2cos2x+1)2+(tanx−3)2=0⇒ cos2x=−12 and tanx=3⇒ 2cos2x−1=−12 and tanx=3⇒ cosx=±12 and tanx=3⇒ x=π3,4π3
Thus, number of solutions of equation is 2.