The original problem is : and is also attached as an image.
I have it to "X=-(cosU * (3^1/2)/2 - sinU * 1/2 ) but that might be wrong, I don't know.
I suppose starting at 2 and then asking for help at 4:51 means I'm going to have to review this chapter, as I've had to ask for help in every section.
I have an interesting method
Consider a cyclic quadralateral and
set is the unit diameter of the circle
Therefore , and
This circle is constructed for this problem
there sum is
we can using the fact that to find the length of ,
Finally , we end up the geometry part by applying Ptolemy's Theorem :
But just slightly flawed, I'm afraid, because it's impossible to draw so that .
The fact is that , and your solution of stops short of solving for . If you take the positive square root and substitute back, you'll find that and . This gives , which represents the solution to the equation .
The negative solution to your equation is, of course, the correct one - and it's the same as the one I obtained by a more conventional approach. Sadly, it's not possible to draw your diagram with a negative value of .
Unless, of course, you can think of a modification ...?
Actually . I didn't realise of the mistake until you told me the impossibility of drawing this ...
Now , i have no confidence to modify my solution because Ptolemy's theorem
requires positve lengths but i am suprised that I still can find out the
exact answer !
Luckily , this method still can be applied to solve this kind of problems if the answer is positive !