How do you solve #1+ cos x + cos 2x = 0#?

Question

Caroline Adams · Accepted Answer

#x=pi/2+kpi,(2pi)/3+2kpi,(4pi)/3+2kpi,kinZZ#

Use the identity for the cosine double angle formula:

#cos2x=2cos^2x-1#

The equation then becomes

#1+cosx+2cos^2x-1=0#

#2cos^2x+cosx=0#

Factor out a #cosx# term.

#cosx(2cosx+1)=0#

Set each of these equal to #0#.

#cosx=0" "" ""and"" "" "2cosx+1=0# #" "" "" "" "" "" "" "" "" "cosx=-1/2#

#cosx=0# occurs at #x=pi/2,(3pi)/2,(5pi)/2#, which can be generalized as #x=pi/2+kpi,kinZZ# (this means where #k# is an integer).

#cosx=-1/2# occurs at #x=(2pi)/3,(4pi)/3# and all other concentric angles (which are #2pi# away), thus we have #x=(2pi)/3+2kpi# and #x=(4pi)/3+2kpi#, both where #kinZZ#.

Nolan Alexander · Answer

undefined To solve the equation $1 + \cos(x) + \cos(2x) = 0$, follow these steps:

1. Use the double-angle identity for cosine: $\cos(2x) = 2\cos^2(x) - 1$.
2. Substitute the double-angle identity into the equation: $1 + \cos(x) + 2\cos^2(x) - 1 = 0$.
3. Simplify the equation: $2\cos^2(x) + \cos(x) = 0$.
4. Factor out common terms: $\cos(x)(2\cos(x) + 1) = 0$.
5. Set each factor equal to zero and solve for $x$:
   - $\cos(x) = 0$ gives solutions where $x$ is an odd multiple of $\frac{\pi}{2}$.
   - $2\cos(x) + 1 = 0$ gives $\cos(x) = -\frac{1}{2}$, which has solutions $x = \frac{2\pi}{3} + 2n\pi$ and $x = \frac{4\pi}{3} + 2n\pi$, where $n$ is an integer.

These are the solutions to the equation $1 + \cos(x) + \cos(2x) = 0$.