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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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