Show that the equation sec θ + cosec θ = c has two roots between 0 and 2π if 2 < c^2 <8 ?

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- August 26th 2008, 07:33 AMfardeen_genShow that the equation sec θ + cosec θ = c has two roots...?
Show that the equation sec θ + cosec θ = c has two roots between 0 and 2π if 2 < c^2 <8 ?

- August 26th 2008, 07:51 AMMoo
Hi,

This one looks easy at first sight (Worried) I think I'll just give you help, not sure for the solution.

(note that c is a real number since the RHS is a real number)

Square the equation :

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Recall that

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The equation is now :

Multiply both sides by (which is , otherwise, the first equation would be nonsense)

. The equation is now :

The discriminant is : , which is__always__positive (because c is a real number).

Solutions are

We know that .

So find the range for c, such that both the solutions are in this interval (Wondering) :

- August 26th 2008, 09:27 AMMoo
Ok, I have a *better* (and more correct) sequence for you (Giggle)

You're asked : prove "******** has 2 roots**if**2<cē<8"

So actually, you just have to prove that

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Write the first inequality :

We want c to satisfy this :

And

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Thus this inequality leads to nothing.

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Write the second inequality :

We want c to satisfy this :

Note that

Therefore

The inequality is now :

and this leads to nothing too... so I'm quite puzzled here (Crying)

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we both don't see any mistake, and the text is correct.... does anyone know what's wrong ? http://smileys.sur-la-toile.com/repo...sol%E9-987.gif - August 26th 2008, 11:56 AMflyingsquirrel
Hello,

I didn't find the mistake your looking for but I've another proof :

Using the graph (see attachment) it seems that for the equation

has two solutions. If we manage to show this then we'll get the result for .

Let for . The derivative of is

As , equals 0 iff that is iff or .

As , is the minimum of on . Similarly is the maximum of on . At this point we've shown that (1) has no solution for if .

Let's show that (1) has exactly one solution in . Let . We have

with and so has same sign as . Thus is increasing on . As is continuous and

and

has at least one solution on this interval (~intermediate value theorem). being increasing, there is exactly one solution.

Similarly one can show that has one solution in hence if , (1) has two solutions.