# Solving Simultaneous Inequalities

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1. Nov 18, 2015

### EternusVia

1. The problem statement, all variables and given/known data

I am trying to solve the simultaneous inequalities (1) and (2) shown in the following image. The solution is provided, but I'm not sure how they solved for it.
http://[ATTACH=full]199971[/ATTACH]
[B]2. Relevant equations[/B]
N/A

[B]3. The attempt at a solution
[/B]
I tried to solve this set of simultaneous inequalities by multiplying (1) by 4, multiplying (2) by -5, and adding them together. My thought was that this would cancel the sin(θ) component and allow me to solve for the range of allowable angles. It did not seem to work.

How does one go about solving simultaneous inequalities?

Last edited by a moderator: May 7, 2017
2. Nov 18, 2015

### Ray Vickson

Last edited by a moderator: May 7, 2017
3. Nov 18, 2015

### EternusVia

Hi Ray,

That is a clever idea. I used it to find part of the solution but couldn't get the rest. Not sure what I'm doing wrong.

$3cos(\theta) + 4sin(\theta) = Asin(\theta + \phi)$
$Asin(\theta + \phi) = Asin(\theta)cos(\phi) + Acos(\theta)sin(\phi)$
$Acos(\phi) = 4$ and $Asin(\phi) = 3$

Thus $tan(\phi) = 3/4$ and $\phi = 36.87, A = 5$

Now we can write inequality (2) and substitute the new expression:

$-4 < 3cos(\theta) + 4sin(\theta) < 2$
$-4 < 5sin(\theta + 36.87) < 2$

Solving, we have $270 < \theta < 346.71$. But from inequality (1), it must also be that $336 < \theta < 53.13$. So one of the solutions is $336 < \theta < 346.71$.

But how do we get the other solution?

*Sorry for the edits

4. Nov 20, 2015

### haruspex

Remember that sin(x)=sin(180-x).

5. Dec 1, 2015

### Shehbaj singh

There's much more easier way which is easy to apply and known by many.
-4<3cosQ + 4sinQ<2
take any one of inequality and have anyone cos or sin to other side and square. when you change one function sin²θ + cos²θ = 1 then you get a quadratic equation. Solve for it you will get intervals where θ lies. Also solve for second inequality in this way and find common intervals which satisfy both equations.
Though this method has some steps more, but it's easy to understand, remember and is very versatile. It could be applied where even there's no trigonometry.

6. Dec 1, 2015

### haruspex

It is not easier, it's more work.
Every time you square an equation you introduce extra (spurious) solutions which have to be weeded out later.

7. Dec 1, 2015

### Shehbaj singh

It will tell you exact intervals. Now if intervals of θ are not joining, there may be two intervals where in between is gap, it will also tell about it. Though, it's some long you can never forget it and it's applicable to many equations. And every solution from it is required and there are no "extra solutions " as you said.QUADRATIC NEVER LIE

8. Dec 1, 2015

### haruspex

Consider the boundary 3cosQ + 4sinQ=2. Your approach, as I understand it, rewrites this as 3cosQ = -4sinQ+2, then squares to produce 9cos2Q = 16sin2Q-16sinQ+4. You then rearrange this into a quadratic in sinQ.
But -3cosQ + 4sinQ=2 leads to the same quadratic, so half the solutions to the quadratic are solutions to that equation. You then have to weed those out.

If you are unconvinced, write out the complete solutions using both methods and post them.