The inequality is
#(x-pi)/(cos^2x)<0#
And the interval is #I=[0,2pi)#
Let
#f(x)=(x-pi)/(cos^2x)#
Build a sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##0##color(white)(aaaaaa)##pi/2##color(white)(aaaaa)##pi##color(white)(aaaa)##3/2pi##color(white)(aaaaa)##2pi#
#color(white)(aaaa)##x-pi##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##cos^2x##color(white)(aaaa)##+##color(white)(aa)##||##color(white)(aa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaa)##-##color(white)(aa)##||##color(white)(aa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
Therefore,
#f(x)<0# when #x in (0, pi/2)uu(pi/2, pi)#
graph{(y-(x-pi)/(cosx)^2)=0 [-16.99, 19.04, -9.44, 8.58]} #