# Thread: Solving for Interval [0, 2pi)

1. ## 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

2. ## 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

3. 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 $cos(2x)=2cos^2(x)-1$

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

This is a quadratic. Let y = cos(x):
$2y^2+2y-1=0$

Using the quadratic formula I get:
$y = \frac{-2 \pm \sqrt{2^2-4*2*-1}}{2*2}$
$y = \frac{-2 \pm \sqrt{12}}{4} = \frac{-1 \pm \sqrt{3}}{2}$

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

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

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

-Dan

4. 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,
$\cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$
Thus,
$\cos 3x+\cos x=0$
Thus,
$2\cos 2x\cos x=0$
Thus,
$\cos 2x\cos x=0$
Thus, set factors equal to zero.
$\left\{ \begin{array}{c} \cos 2x=0\\ \cos x=0$

5. You already asked that question in "Urget Math". DO NOT MAKE DOUBLE POSTS
-=USER WARNED=-

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