Trigo Identities!

• February 19th 2010, 12:42 AM
gp3
Trigo 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!
• February 19th 2010, 02:06 AM
Hello gp3
Quote:

Originally Posted by gp3
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!

Use the identity:
$\cos 2x = 2\cos^2x-1$
Then simplify the LHS to get a quadratic in $\cos x$, which you can then factorise.

Can you do it now?

• February 19th 2010, 02:39 AM
gp3
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)
• February 19th 2010, 02:45 AM
e^(i*pi)
Quote:

Originally Posted by gp3
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)

No, you've missed out your 3cos(x) term.
FYI $2cos^2(x)+1=0$ has no real solutions. This is because $x^2 \geq 0$ for all real x. Adding 1 will always result in a positive answer where x is real

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

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

I have used square brackets to show this more effectively

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

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

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

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

$(2\cos (x) -1)(\cos (x) - 1) = 0$
• February 19th 2010, 02:49 AM
gp3
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 ?
• February 19th 2010, 04:06 AM
Punch
Quote:

Originally Posted by gp3
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 ?

Just solve it like any other quadratic equation. Treat cosx as x.