Question

How do I solve for the angle when cos2x+cosx=0?
havent done in a while need refreshing

Answer 1

No one? really need it

Answer 2

I think it may be no solution?

Answer 3

There is one; I'm working on it

Answer 4

okay thanks

Answer 5

If cos(x) + cos(2x) = 0, cos(x) = -cos(2x)

We see 3 possible cases here.

Since cos(k*pi) = 1 for even k and -1 for odd k, one solution is:
x = pi + k*2*pi. Since 2x = 2*pi + 4*k*pi, one x will have an odd coefficient and 2x will have an even one, so cos(x) = -1 and cos(2x) = 1.

The other two cases are when x = 1/3*pi + 2*pi*n (since cos(1/3*pi) = cos(2/3*pi) and adding 2*pi*n to 1/3*pi adds 4*pi*n to 2/3*pi, so the locations of the angles will be the same with).

We see the exact same thing for -1/3*pi, so the other set of solutions is x = 2*pi*n - 1/3pi

Thus:

x = pi + 2*pi*n
x = 1/3*pi + 2*pi*n
x = -1/3*pi + 2*pi*n