Number of solutions of the equation 4cos2⁡2x+cos⁡2x+1+tan⁡x(tan⁡x−23)=0 in [0, 2π] is

4cos2⁡2x+4cos⁡2x+1+tan2⁡x−23tan⁡x+3=0⇒ (2cos⁡2x+1)2+(tan⁡x−3)2=0⇒ cos⁡2x=−12 and tan⁡x=3⇒ 2cos2⁡x−1=−12 and tan⁡x=3⇒ cos⁡x=±12 and tan⁡x=3⇒ x=π3,4π3∴ Number of solutions of equation = 2