|
I just want to see who of the knowledge people will solve this first. Time yourself
Solve the equation: 2 cos(sq) x + 6 cos x - 3 = 0. Restrict solutions to the interval [0, 2pi.]
|
i'm not going to do your homework for you, but i can tell you that you can only solve this problem numerically, not analytically.
|
actually, this isn't my homework and I have the answer already.
I don't even have trig as a class. So, any other takers? :) |
what is cos(sq)?
|
What's the point of solving this? Just to demonstrate if we know trig, or what? I took trig many many years ago and could solve this one if I sat down to do it - but what's the point?
Solving trig functions lost it's joy for me quite a long time ago ;) |
2 cos(sq). The "sq" is just my way of writing "squared"
Sorry, should have specified before. My mistake :) |
You can solve it analytically. Set y=cos(x), then use the quadratic formula. Then apply arccos(y).
|
I used u, but phukraut is right, whatever letter you use.
|
Quote:
A=cos x 2A^2+6A-3=0 The quadratic cannot be factored so we use the quadratic equation: A = (-3+/- sqrt 15)/2 the negative will give A<-1 which is impossible for A = cos x. cos x = (-3 + sqrt15)/2 Not solvable anaytically. Numerically, x = 1.1191.... |
cos(x) = (-6 +- sqrt(60))/4
= 0.436491673, -3.43649... Obviously, cos(x) is between -1 and 1 so the second root is not true. cos^-1(0.436491673) = 1.119100764, 5.164084543 I really hope that's not your homework. Edit, must of been playing around with it whilst Amano was posting... |
According to <a href="http://mathworld.wolfram.com/Analytic.html">MathWorld</a>,
Quote:
|
Very nice, Amano. If cos x = 0.4365 the calculator gives x=1.1191 as the acute solution.
And we deduce that the other solution is x = 2pi - 0.4365 = 5.1641. The equation of cos x = -3.4365 has no solutions, since cos x < 1 for all x. The solution set is therefore {0.4365, 5.1641}. - |
ok so here's another one:
Find sin (pi/5). And no fair using google! |
hmm, (2pi over 5) = 5 + 2 radical 5
I wish there were the actual symbols to put on here, because I can't ever remember the correct names |
| All times are GMT -8. The time now is 08:22 AM. |
Powered by vBulletin® Version 3.8.7
Copyright ©2000 - 2025, vBulletin Solutions, Inc.
Search Engine Optimization by vBSEO 3.6.0 PL2
© 2002-2012 Tilted Forum Project