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.

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....

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>,

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

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