Solving $\cos(2\theta)=\cos(\theta)$ The question :
Solve Equation $\cos(2\theta)=\cos(\theta)$ for $0 \le \theta \le 2\pi$ in terms of $\pi$.
My solution: -
Using trigonometry identity , I quickly get
$$2\cos^2(\theta)-1=\cos(\theta)$$replcae $$\cos(\theta)=a.$$
I'll get quadratic equation of
$$2a^2-a-1=0,$$
where $a$ is $$a = -\frac12 , 1.$$
The result I got is$$\theta=0,{2\over3}\pi,2\pi.$$
The scheme said I'm missing is $\frac43\pi$.
Where am I missing? How to achieve it?
 A: The first thing missing in your solution is the '$^2$' on the LHS of your equation, as @Gary has spotted out.

$$2cos(\theta)-1=cos(\theta)$$

It should be $$2\cos^2(\theta)-1=\cos(\theta).$$
Then you've let $a = \cos\theta$ and corrected factorized it to $(2a+1)(a-1) = 0$, so $a = -\dfrac12$ or $a = 1$.
Here's a more visual interpretation of @Semiclassical and @Christoph's comments.
Note that $\cos \theta$ is the $x$-coordinate of the green point on the unit circle.  The vertical line $x = -\dfrac12$ intersects the unit circle at two points (symmetric about the $x$-axis), giving solutions $\theta = \pi \pm \dfrac\pi3 = \dfrac{2\pi}3$ or $\dfrac{4\pi}3$.
Even though the vertical line $x = 1$ touches the unit circle at only one point $(1,0)$, as $\theta\in[0,2\pi]$, we have $\theta = 0$ or $2\pi$.

Image source: Khan Academy
A: $\cos (2\pi-\frac {2 \pi} 3)=\cos \frac {2 \pi} 3=-\frac 1 2$ and $2\pi-\frac {2\pi} 3=\frac {4\pi} 3$.
A: HINT: When $cos(\theta) = \frac{-1}{2}$ which means  $\theta = \pm\frac{2}{3} \pi + 2k\pi \ \ k \in \mathbb{Z}$
A: $$\cos\theta  =c,  \cos 2 \theta =2 c^2-1, \; 2c^2-c-1= (c-1) (2c+1)=0,\quad c= (1, -\dfrac12);\; \theta= (0,2\pi/3,4\pi/3)..$$
and the cyclic co-terminal angles.
