Solve for x on #[0,2pi)# : #(x-pi)/cos^2x < 0 # ?

2 Answers
Jul 17, 2018

#x< pi# and #xne pi/2#

Explanation:

It must be #cos^2(x)ne 0# since #cos^2(x)# is the denominator of the given inequality and #cos^2(x)>0# in the given interval except #x=pi/2#.
So we get #x < pi#

Jul 17, 2018

The solution is #x in (0, pi/2)uu(pi/2, pi) #

Explanation:

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]} #