The number of roots of the equation,

cot x=π/2+x in [-π,3π/2] is,

hi, put x = cot^-1(#) so, it becomes # = pie/2 + cot^-1(#) & #-pie/2 = cot^-1(#) # = -cot{pie/2 - #} implies # = -tan(#) & from above # = cotx implies cotx = -tan(cotx) in above domain this eqn is satisfied by x = {-pie/2, pie/2, 3pie/2} so no. of roots are 3.