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Need help solving a trigonomic equation!

  • #1
I hope I'm posting this in the correct spot?!

I just got hired a few months ago as an mechanical engineer at a small start up company and need help solving an equation that I've spent hours after hours on trying to figure out. If anyone could offer advice and or steps to get the solution I'd very much appreciate it. I think I've tried almost everything but apparantly not cause I can't figure out how to solve for x. Here is the equation: 0.5*cos(x)-0.875*sin(x)*cos(x)+A*sin^2(x)=A

I'd show you all the steps I've tried but it would take me several hours to type them all in here. I tried converting everything to both sin and cos but that didn't seem to work. Thanks in advance to any geniouses out there...
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,989
6
Notice that if you collect terms with A, you get [itex]A(1-sin(x)^2)=Acos^2(x)[/itex], since sin^2(x)+cos^2(x)=1

Then you have
[tex](\frac{1}{2}-\frac{7}{8}\sin(x))\cos(x)=A\cos^2(x)[/tex]

So all x for which cos(x)=0 are solutions.
To find other solutions, assume cos(x) is not 0 and divide both sides by cos(x). You can take it from there.
 
  • #3
Galileo said:
Notice that if you collect terms with A, you get [itex]A(1-sin(x)^2)=Acos^2(x)[/itex], since sin^2(x)+cos^2(x)=1

Then you have
[tex](\frac{1}{2}-\frac{7}{8}\sin(x))\cos(x)=A\cos^2(x)[/tex]

So all x for which cos(x)=0 are solutions.
To find other solutions, assume cos(x) is not 0 and divide both sides by cos(x). You can take it from there.
Thank you so much, that helps! Can't believe I didn't see that. But I'm still struggling... I end up getting this equations which I can't figure out how to solve (it's been a few years...):
4=7*sin(x)+8*A*cos(x)
I'm probably missing something very simple here I'm guessing?
 
  • #4
Galileo
Science Advisor
Homework Helper
1,989
6
A linear combination of a sine and a cosine is again a sinusoid with the same period. I suppose you could use:
[tex]a\cos(x)+b\sin(x)=\sqrt{a^2+b^2}\cos(x-\phi)[/tex]
where
[tex]\phi=\arctan(b/a)[/tex]

It looks nice, but I`m afraid it will be messy.
 
  • #5
Thank you so much Galileo. You truely helped me out a ton and I was able to find the solution with your help. Rock on. Case closed.
 

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