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Trig Inequality

  • Thread starter dorkee
  • Start date
1. The problem statement, all variables and given/known data
Given 0≤x<2pi, solve tan x/2 > -1



3. The attempt at a solution
I thought I would set tan x/2=-1 but I'm not sure.

Is the answer 0<x<pi, 3pi/2<x<2pi?
 
Last edited:
Your answer doesn't look right to me.

Why not try a graphical approach? From 0 to 2pi there will be 4 periods, and it looks like there will be 4 intervals in your solution.

Oops! The period is 2pi, not pi/2. Yes, it seems your solution is right. Without a graph, I like your approach of solving the equality. Then plug in some test points on either side of each solution you get. Pay attention to asymptotes (x = pi in this case).
 

HallsofIvy

Science Advisor
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1. The problem statement, all variables and given/known data
Given 0≤x<2pi, solve tan x/2 > -1



3. The attempt at a solution
I thought I would set tan x/2=-1 but I'm not sure.

Is the answer 0<x<pi, 3pi/2<x<2pi?
Yes, solving tan(x/2)= -1 is the way to start. On any interval NOT including a root of tan(x/2)= -1 and NOT including [itex]\pi[/itex], where tan(x/2) is not continuous, tan(x/2) is always less than -1 or always less than -1. Determine what those intervals are and check one point in each interval to see if tan(x/2), on that interval, is greater than -1.
 
I got one question.

What if I have cosx > -1 and 0≤x<2п ?

x Є (-п, п) or x Є (п,3п) оr x Є (п,п+2kп), k Є Z

But because of 0≤x<2п would x Є (0,п) or x Є (п,2п) or x Є ( kп, п(k+1) ) where k Є Z.

Am I right?
 
Your first answer is closer than your second answer. If you're restricted to [0, 2pi) then there's no need to mess with the k values.

Two hints: cos(0) = 1 and cos(pi) = -1
 
Your first answer is closer than your second answer. If you're restricted to [0, 2pi) then there's no need to mess with the k values.

Two hints: cos(0) = 1 and cos(pi) = -1
Yes, you're right. I don't want to mess with k values.

So my answer is (0,п) or (п,2п)

And what about cosx < -1 . As I know there isn't smallest value than -1 in cos. So is x Є empty set?

Thanks for the help.

Regards.
 

HallsofIvy

Science Advisor
41,625
822
Yes, {x| cos(x)< -1} is the empty set.
 
Right. And remember that, for the original problem, zero is part of the solution. So your interval should start with a [ instead of a (
 

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