My question states: cos(3x)+cos(x)=0

I got two answer pi/2 and pi/4 but there are supposed to be more, can you show me how to get all the answers?

Thanks alot

Printable View

- May 8th 2006, 01:06 PMaussiekid90Trig Product Formula Problem
My question states: cos(3x)+cos(x)=0

I got two answer pi/2 and pi/4 but there are supposed to be more, can you show me how to get all the answers?

Thanks alot - May 8th 2006, 01:12 PMaussiekid90Solving for Interval [0, 2pi)
This problem was pretty tough for me:

cos(2x)+2cos(x)=0

can you show me how to do this? Thanks!

-Aussiekid90 - May 8th 2006, 01:33 PMtopsquarkQuote:

Originally Posted by**aussiekid90**

Thus

$\displaystyle cos(2x)+2cos(x)=0$

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

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

This is a quadratic. Let y = cos(x):

$\displaystyle 2y^2+2y-1=0$

Using the quadratic formula I get:

$\displaystyle y = \frac{-2 \pm \sqrt{2^2-4*2*-1}}{2*2}$

$\displaystyle y = \frac{-2 \pm \sqrt{12}}{4} = \frac{-1 \pm \sqrt{3}}{2}$

Thus $\displaystyle cos(x) = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}$

Now. $\displaystyle -\frac{1}{2} - \frac{\sqrt{3}}{2} < -1$ so we need to take the positive sign. As it happens the result doesn't seem to have a nice form, so:

$\displaystyle x = cos^{-1} \left ( -\frac{1}{2} + \frac{\sqrt{3}}{2} \right ) \approx 1.19606 \, rad$. Since this angle is second quadrant, we also get an angle $\displaystyle x = 2 \pi \, rad - 1.19606 \, rad = 5.08712 \, rad$.

-Dan - May 8th 2006, 01:47 PMThePerfectHackerQuote:

Originally Posted by**aussiekid90**

$\displaystyle \cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right) $

Thus,

$\displaystyle \cos 3x+\cos x=0$

Thus,

$\displaystyle 2\cos 2x\cos x=0$

Thus,

$\displaystyle \cos 2x\cos x=0$

Thus, set factors equal to zero.

$\displaystyle \left\{ \begin{array}{c} \cos 2x=0\\ \cos x=0$ - May 8th 2006, 01:58 PMThePerfectHacker
You already asked that question in "Urget Math". DO NOT MAKE DOUBLE POSTS

-=USER WARNED=-

The thread has been merged. - May 8th 2006, 02:07 PMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

cos(2x)+2cos(x)=0

and the other is:

cos(3x)+cos(x)=0

They're close, but not quite the same.

-Dan