Help on a problem, involves trig and possible trig identities.

cos(2x) = -cos(x)

2cos^(x)  - 1=  - cos(x)

2cos^2(x) + cos(x) - 1   = 0     factor

(2cos(x) - 1) (cos(x) + 1)  =  0     and setting each factor to 0, we have

2cos(x) - 1 = 0                                                                      

add 1 to both sides                                                               

2cos(x)  = 1                                                                            

divide both sides by 2                                                              

cos(x) = 1/2                                                                            

And this happens at   pi/3 + 2pi(n)

and at   5pi/3 + 2pi(n)

where n is an integer

 cos(x) + 1 =  0 

 subtract 1 from each side

 cos(x)  = -1

 and this happens at pi + 2pi(n)

where n is some integer

Here's a graph of the solutions........https://www.desmos.com/calculator/kietyhmst0

How do i solve, cos(2x)=-cos(x)

$\small{ \begin{array}{lrcl} & \cos{(2x)} &=& -\cos{(x)} \\ & \mathbf{\cos{(2x)} + \cos{(x)} } & \mathbf{=} & \mathbf{0} \\ \\ \hline \\ \text{Formula derivation } & \cos{(ux+vx)} = \cos{(ux)}\cos{(vx)} - \sin{(ux)}\sin{(vx)}\\ & \cos{(ux-vx)} = \cos{(ux)}\cos{(vx)} + \sin{(ux)}\sin{(vx)}\\ & \cos{(ux+vx)} + \cos{(ux-vx)} = 2 \cos{(ux)}\cos{(vx)} \\ & & ux+vx = nx &\qquad uv-vx = mx\\ & &(1)~~ u+v = n & \qquad (2)~~u-v = m\\ (1) + (2) & n+ m = 2u \qquad u=\frac{n+m}{2}\\ (1) - (2) & n- m = 2v \qquad v=\frac{n-m}{2}\\ \end{array} }\\ \small{ \boxed{~ \cos{(nx)} + \cos{(mx)} = 2 \cos{ \left(\frac{n+m}{2}x \right)}\cos{ \left(\frac{n-m}{2}x \right)} ~} }$

$n=2 \qquad m=1\\ u=\frac{2+1}{2} = \frac32\\ v= \frac{2-1}{2}=\frac12\\ $

$\small{ \begin{array}{rcl} \cos{(2x)} + \cos{(x)} &=& 2\cos{ \left(\frac32 x \right) }\cos{ \left(\frac12 x \right) } &=& 0\\ 2\cos{ \left(\frac32 x \right) }\cos{ \left(\frac12 x \right) } &=& 0 \qquad | \qquad : 2 \\ \cos{ \left(\frac32 x \right) }\cos{ \left(\frac12 x \right) } &=& 0 \qquad \text{ setting each factor to 0}\\ \end{array} }\\ \small{ \begin{array}{lrcll} \\ 1. & \cos{ \left(\frac32 x \right) } &=& 0 \qquad & | \qquad \pm\arccos{()} \\ & \frac32 x &=& \pm\arccos{(0)} \pm 2k\pi \qquad & | \qquad \arccos{(0)} = \frac{\pi}{2} \\ & x &=& \frac{2}{3} ( \pm\frac{\pi}{2} \pm 2k\pi )\\ &\mathbf{ x } & \mathbf {=} & \mathbf{ \pm\frac{\pi}{3} \pm \frac{4}{3}k\pi }\qquad & | \qquad k \in N \\ \end{array} }\\ \small{ \begin{array}{lrcll} \\ 2. & \cos{ \left(\frac12 x \right) } &=& 0 \qquad & | \qquad \pm\arccos{()} \\ & \frac12 x &=& \pm\arccos{(0)} \pm 2k\pi \qquad & | \qquad \arccos{(0)} = \frac{\pi}{2} \\ & x &=& 2 \cdot ( \pm\frac{\pi}{2} \pm 2k\pi )\\ &\mathbf{ x } & \mathbf {=} & \mathbf{ \pm\pi \pm 4\cdot k\pi }\qquad & | \qquad k \in N \\ \end{array} }\\$