Show that the equation sec θ + cosec θ = c has two roots between 0 and 2π if 2 < c^2 <8 ?
Hi,
This one looks easy at first sight 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 :
Ok, I have a *better* (and more correct) sequence for you
You're asked : prove "******** has 2 roots if 2<c²<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
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we both don't see any mistake, and the text is correct.... does anyone know what's wrong ?
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.