The given equation is defined for x=2π,23π.Now, tanx+secx=2cosx⇒cosxsinx+cosx1=2cosx⇒(sinx+1)=2cos2x⇒(sinx+1)=2(1−sin2x)⇒(sinx+1)=2(1−sinx)(1+sinx)⇒(1+sinx)[2(1−sinx)−1]=0⇒2(1−sinx)−1=0{[∵sinx=−1 otherwise cosx=0 and tanxsecx will be undefined ]}⇒sinx=21⇒x=6π,65π in (0,2π)