Given, equation is tanx+secx=2cosx ⇒cosxsinx+cosx1=2cosx ⇒1+sinx=2cos2x ⇒1+sinx=2(1−sin2x) ⇒2sin2x+2sinx−sinx−1=0 ⇒2sinx(sinx+1)−1(sinx+1)=0 ⇒(2sinx−1)(sinx+1)=0 ⇒ either sinx=21 or sinx=−1 ⇒ either x=6π,65π∈[0,π] or x=23π
But, x=23π can not be possible. ∴ Number of solutions are 2.