I just want to see who of the knowledge people will solve this first. Time yourself - Tilted Forum Project Discussion Community

Cryptic · 11-19-2004, 02:55 PM

Solve the equation: 2 cos(sq) x + 6 cos x - 3 = 0. Restrict solutions to the interval [0, 2pi.]

Amano · 11-19-2004, 03:06 PM

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.

Cryptic · 11-19-2004, 03:13 PM

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?

phukraut · 11-19-2004, 03:19 PM

what is cos(sq)?

Pragma · 11-19-2004, 03:20 PM

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

Cryptic · 11-19-2004, 03:20 PM

2 cos(sq). The "sq" is just my way of writing "squared"
Sorry, should have specified before. My mistake

phukraut · 11-19-2004, 03:24 PM

You can solve it analytically. Set y=cos(x), then use the quadratic formula. Then apply arccos(y).

Cryptic · 11-19-2004, 03:27 PM

I used u, but phukraut is right, whatever letter you use.

Amano · 11-19-2004, 03:29 PM

Quote:

Originally Posted by Cryptic

actually, this isn't my homework

very well then.
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....

molloby · 11-19-2004, 03:35 PM

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...

phukraut · 11-19-2004, 03:36 PM

According to <a href="http://mathworld.wolfram.com/Analytic.html">MathWorld</a>,

Quote:

A solution to a problem that can be written in "closed form" in terms of known functions, constants, etc., is often called an analytic solution.

I would consider arccos a known function. Maybe that's up for debate.

Cryptic · 11-19-2004, 03:37 PM

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}. -

Amano · 11-19-2004, 03:45 PM

ok so here's another one:

Find sin (pi/5).

And no fair using google!

Cryptic · 11-19-2004, 03:53 PM

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

11-19-2004, 02:55 PM	#1 (permalink)
Cryptic Upright	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.] __________________ F=MA 2.998*108ms-1 Ek = 1/2mv2 R D R R

11-19-2004, 03:13 PM	#3 (permalink)
Cryptic Upright	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? __________________ F=MA 2.998*108ms-1 Ek = 1/2mv2 R D R R

11-19-2004, 03:20 PM	#6 (permalink)
Cryptic Upright	2 cos(sq). The "sq" is just my way of writing "squared" Sorry, should have specified before. My mistake __________________ F=MA 2.998*108ms-1 Ek = 1/2mv2 R D R R

11-19-2004, 03:27 PM	#8 (permalink)
Cryptic Upright	I used u, but phukraut is right, whatever letter you use. __________________ F=MA 2.998*108ms-1 Ek = 1/2mv2 R D R R

11-19-2004, 03:35 PM	#10 (permalink)
molloby Tilted Location: Sydney, Australia	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... Last edited by molloby; 11-19-2004 at 03:37 PM..

11-19-2004, 03:06 PM	#2 (permalink)
Amano Insane	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.

11-19-2004, 03:19 PM	#4 (permalink)
phukraut Addict	what is cos(sq)?

11-19-2004, 03:20 PM	#5 (permalink)
Pragma I am Winter Born Location: Alexandria, VA	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

11-19-2004, 03:24 PM	#7 (permalink)
phukraut Addict	You can solve it analytically. Set y=cos(x), then use the quadratic formula. Then apply arccos(y).

11-19-2004, 03:37 PM	#12 (permalink)
Cryptic Upright	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}. - __________________ F=MA 2.998*108ms-1 Ek = 1/2mv2 R D R R

11-19-2004, 03:45 PM	#13 (permalink)
Amano Insane	ok so here's another one: Find sin (pi/5). And no fair using google!

11-19-2004, 03:53 PM	#14 (permalink)
Cryptic Upright	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 __________________ F=MA 2.998*108ms-1 Ek = 1/2mv2 R D R R