1. ## inequalities

By means of graph, solve the inequality
tan2x+1≤4

I’ve tried like (2tanx)/(1-tan2) -3≤0
then (3tan2x+2tanx-3)/(1+tanx)(1-tanx) ≤0

can anyone point out my mistake?

2. ## Re: inequalities

Have you actually graphed it?

What does the graph tell you? (The point is that you are not going to find a closed form solution so you estimate the solutions via graphing.)

-Dan

3. ## Re: inequalities

Is that tan(2x)+ 1 or tan(2x+ 1)?

4. ## Re: inequalities

Originally Posted by HallsofIvy
Is that tan(2x)+ 1 or tan(2x+ 1)?
tan(2x) +1

5. ## Re: inequalities

If you have properly described the question, you are to solve this BY MEANS OF A GRAPH.

As topsquark pointed out, this requires sketching the pertinent graph and no analysis at all.

To deal with this analytically rather than geometrically, I'd probably start with u = 2x which reduces the problem to $tan(u) \le 3.$

Then I'd use the inverse trig functions.

6. ## Re: inequalities

Originally Posted by JeffM
If you have properly described the question, you are to solve this BY MEANS OF A GRAPH.

As topsquark pointed out, this requires sketching the pertinent graph and no analysis at all.

To deal with this analytically rather than geometrically, I'd probably start with u = 2x which reduces the problem to $tan(u) \le 3.$

Then I'd use the inverse trig functions.
I understand that. And I've sketched the graph, but I'm not sure still..

let tan2x=3
2x=71.57, 251.57, 431.57, 611.57,
x=35.79, 125.79, 215.79, 305.79
so, 0≤x≤35.79, 45<x≤125.79, 135<x≤215.79, 225<x≤305.79, 315<x≤360??

7. ## Re: inequalities

Originally Posted by Trefoil2727
I understand that. And I've sketched the graph, but I'm not sure still..

let tan2x=3
2x=71.57, 251.57, 431.57, 611.57,
x=35.79, 125.79, 215.79, 305.79
so, 0≤x≤35.79, 45<x≤125.79, 135<x≤215.79, 225<x≤305.79, 315<x≤360??
How you answer depends on exactly how the question is asked and what you have been taught.

Are you to answer in radians or degrees? Are there any additional limits placed on x? Do you know how to represent a cyclic answer?

$cos( \theta ) = 0 \implies \theta = \dfrac{\pi }{2} * (2k - 1)\ for\ all\ k \in \mathbb Z.$

8. ## Re: inequalities

[QUOTE=JeffM;823365]How you answer depends on exactly how the question is asked and what you have been taught.

Are you to answer in radians or degrees? Are there any additional limits placed on x? Do you know how to represent a cyclic answer?

9. ## Re: inequalities

$tan(2x) + 1 \le 4 \implies tan(2x) \le 3.$

$0 < v < 90^o\ and\ v = arctan(3) \implies v \approx 71.5650^o < 90^o.$

$-90^o < 2x < 90^o\ and\ tan(2x) \le 3 \implies -45^o < x\ and\ 2x \le v \implies - 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$

So one possible answer is $- 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$

A better answer might be $\{-45 + (2k - 1) * 90\}^o < x \le \left\{\dfrac{arcsin(3)}{2} + (2k - 1) * 90\right\}^o,\ for\ all\ k \in \mathbb Z.$

10. ## Re: inequalities

Originally Posted by JeffM
$tan(2x) + 1 \le 4 \implies tan(2x) \le 3.$

$0 < v < 90^o\ and\ v = arctan(3) \implies v \approx 71.5650^o < 90^o.$

$-90^o < 2x < 90^o\ and\ tan(2x) \le 3 \implies -45^o < x\ and\ 2x \le v \implies - 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$

So one possible answer is $- 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$

A better answer might be $\{-45 + (2k - 1) * 90\}^o < x \le \left\{\dfrac{arcsin(3)}{2} + (2k - 1) * 90\right\}^o,\ for\ all\ k \in \mathbb Z.$
Why are you using 2k-1? There is no possible value of $k$ such that you get the inequality you gave (second line from bottom of your quote).

11. ## Re: inequalities

Ooops. Should have been 2k, not 2k - 1. No clue what cog slipped there.