If  2tan2x−5secx=1 for exactly 10 distinct values of  x∈0,nπ,n∈N  then the greatest value of n is

2sec2x−1−5secx=1        ⇒2sec2x−5secx−3=0⇒secx−32secx+1=0⇒secx=3⇒cosx=13Which gives two values of x in each of  0,2π,(2π,4π],(4π,6π],(6π,8π] and two values in   (8π,10π]      ∴   greatest value of n = 10.