Question

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

Answer 1

`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`

Answer 2

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]} #