Assume tan2 x = a where a ' belongs to 'R
now , equation becomes 5 a 2 - 4 a - 1 = 0
factorise and you'll get ( 5 a + 1 ) ( a - 1 ) = 0
= 5 a = -1 or a = 1
= a = ( - 1 / 5 ) or a =1
Since ,square of any function ( even negative numbers ) is never negative
so , a doesn't equal to ( - 1 / 5 ) and hence , a = 1 is true
= tan x = ± 1
this gives the general solution as { n ± ( / 4 ) } , where n ' belongs to ' Z