triangle and trig: x^2 - 2^(5/4)x - (2^(3/2) - sinx - cosx)

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x^2 - 2^(5/4)x - (2^(3/2) - sinx - cosx) = 0;
have two roots p & q.
A triangle contain angle w and two line with length p & q.
Find w and the area of triangle!
 
Is w the angle formed by the intersection of sides p and q? Does that matter?

I'm thinking DesCarte's Rule of Signs on this one. Let's just play like it's a polynomial.

\(\displaystyle sin(x)+cos(x) = \sqrt{2}*sin(x+\frac{\pi}{4})\)

This gives the "constant"-ish term as \(\displaystyle \sqrt{2}*(2-sin(x+\frac{\pi}{4}))\).

That expression is easily demonstrated never to be negative.

I think we're done. Either p or q must be less than zero. No such triangle.
 
You don't actually need that result. It's just more obvious if you have it. It's not necessarily part of your problem, so I wouldn't worry too much about it.

You have 2^(3/2) - sinx - cosx.

The important part is that 2^(3/2) > 2 and both sin(x) and cos(x) are at most 1.

(Greater than 2) - 2*(At most 1) is ALWAYS greater than 0. Same result.

You would have a problem if that constant were hanging around somewhere in the zone \(\displaystyle \sqrt{2} < Constant < 2\). If that were the case, you would need that result that I told you not to worry about. If the constant happened to be less than \(\displaystyle \sqrt{2}\), then there might be a triangle worth talking about.
 
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