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This suddenly felt impossible:
How do I transform these three equations...
(1): xz=bc
(2): cos(y) + xsin(y)=0
(3): zsin(y) + xzcos(y) - c(a+b)=0
(three unknown x,y,z with three equations, always solvable right!? But i only want to solve this for x!!!)
(I've looked it through carefully, so it shouldn't be any errors in there)
...to either of these too:
answer, version1: x=1/√((a/b+1)²-1)
, version2: x=tan(arcsin(b/(a+b))
This is how far I've got:
(1) can be rewritten to: x=bc/z , new(4)
(2) -"- : x=cos(y)/sin(y) , new(5)
(3) -"- : z=c(a+b)/(sin(y)+xcos(y)) ,new(6)
(6) in (4) --> x=b(sin(y)+xcos(y))/(a+b) , new(7)
(5) in (7) --> cos(y)/sin(y)=b(sin(y)+cos²(y)/sin(y))/(a+b) --> [multiplying both sides with sin(y)] cos(y)=b/(b+c)
So, finally: I'm getting y=arccos(b/(a+b)), BUT HOW DO I PROCEED... It's a dead end. I don't know what more to do...
Maybe you do?
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CORRECTION: eq(2) should be: cos(y) - xsin(y)=0!
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(three unknown x,y,z with three equations, always solvable right!? But i only want to solve this for x!!!)
There are actually six variables...x,y,z,a,b,c
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Yes, but he can measure a, b and c, so he wants to make x the subject of an equation involving just those.
Why did the vector cross the road?
It wanted to be normal.
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