# Solving for Interval [0, 2pi)

• May 8th 2006, 01:06 PM
aussiekid90
Trig 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 PM
aussiekid90
Solving 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 PM
topsquark
Quote:

Originally Posted by aussiekid90
This problem was pretty tough for me:

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

can you show me how to do this? Thanks!

-Aussiekid90

Recall that $\displaystyle cos(2x)=2cos^2(x)-1$

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 PM
ThePerfectHacker
Quote:

Originally Posted by aussiekid90
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

Use identity,
$\displaystyle \cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$
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 PM
ThePerfectHacker
You already asked that question in "Urget Math". DO NOT MAKE DOUBLE POSTS
-=USER WARNED=-
• May 8th 2006, 02:07 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
You already asked that question in "Urget Math". DO NOT MAKE DOUBLE POSTS
-=USER WARNED=-