hi guys,

solve cos2x - 3cosx +2 = 0 for 0 < x < 360

which formula should I use? is it the sum to product? I dont really know how to go about doing it! some advice pls!

thanks!

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- Feb 18th 2010, 11:42 PMgp3Trigo Identities!
hi guys,

solve cos2x - 3cosx +2 = 0 for 0 < x < 360

which formula should I use? is it the sum to product? I dont really know how to go about doing it! some advice pls!

thanks! - Feb 19th 2010, 01:06 AMGrandad
- Feb 19th 2010, 01:39 AMgp3
so what I've got now is:

2cos square x - 1 + 2 = 0

2cos square x + 1 = 0

(cosx+1)(cosx+1) = 0

is this correct? (Surprised) - Feb 19th 2010, 01:45 AMe^(i*pi)
No, you've missed out your 3cos(x) term.

FYI $\displaystyle 2cos^2(x)+1=0$ has no real solutions. This is because $\displaystyle x^2 \geq 0$ for all real x. Adding 1 will always result in a positive answer where x is real

The original question is $\displaystyle \cos(2x) - 3\cos(x) +2 = 0 $

What Grandad said is to change $\displaystyle \cos(2x) to 2\cos^2(x)-1$ - the 3cos(x) term is unaffacted.

I have used square brackets to show this more effectively

$\displaystyle [2\cos^2(x) - 1] - 3cos(x) + 2 = 0$

You can simplify the constant terms to get it into the form $\displaystyle ax^2+bx+c=0$:

$\displaystyle 2\cos^2(x) - 3cos(x) + 1 = 0$

You can the solve the quadratic using your favourite method although this one factorises.

$\displaystyle (2\cos (x) -1)(\cos (x) - 1) = 0$ - Feb 19th 2010, 01:49 AMgp3
oh yea, thanks alot, I missed out 3cosx, guess I was too in a hurry.

and btw, just a quick one, how do you factorise the eqn, from 2cos square x -3 cosx+1+0 to (2 cosx-1)(cosx-1)= 0 ? - Feb 19th 2010, 03:06 AMPunch