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
Recall that $\displaystyle cos(2x)=2cos^2(x)-1$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
Use identity,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.
$\displaystyle \left\{ \begin{array}{c} \cos 2x=0\\ \cos x=0$