Solving Simultaneous Inequalities

[EternusVia]

Homework Statement

[/B]
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]
[h2]Homework Equations[/h2]
N/A

[h2]The Attempt at a Solution[/h2]
[/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?

 

[Ray Vickson]

Homework Statement

[/B]
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]199972[/ATTACH]
[h2]Homework Equations[/h2]
N/A

[h2]The Attempt at a Solution[/h2]
[/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?[/QUOTE]

Write $3 \cos \theta + 4 \sin \theta$ as $A \sin(\theta + \phi)$ for known values of $A,\phi$. This is done using trigonometric identities $A \sin(\theta + \phi) = A \sin(\theta) \cos(\phi) + A \cos(\theta) \sin(\phi)$, so you need to solve $A \cos \phi = 4, A \sin \phi = 3$. That is easy enough to do if you use additional trigonometric identities.

Anyway, from the two equations you will get $a_1 < \theta < a_2$ and $b_1 < \theta + \phi < b_2$, so then you just need to figure out the common $\theta$-region.

 

[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

 

[haruspex]

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

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

 

[Shehbaj singh]

There's much more easier way which is easy to apply and known by many.
-4<3cosQ + 4sinQ<2
take anyone 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.

 

[haruspex]

There's much more easier way which is easy to apply and known by many.
-4<3cosQ + 4sinQ<2
take anyone 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.

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.

 

[Shehbaj singh]

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.

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

 

[haruspex]

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

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.