What is the general solution to $2 \cos^2 x-\cos x=0$? What is the general solution to this trig equation?
$$2 \cos^2 x-\cos x=0$$
Thanks.
 A: Notice that $$2 \cos^2 x-\cos x=0\implies 2\cos^2x=\cos x$$
So either $\cos x=0$ or $\cos x=\frac{1}{2}$. 
Now find general solution
A: $$2\cos^2(x) - \cos(x) = 0 \implies \cos(x)(2\cos(x) - 1) = 0$$
For what $x$ is $\cos(x) = 0, \frac{1}{2}$? There are only 2 solutions that form all distinct solutions; all others are $2k\pi{}$ away from them
Have a look
A: $2\cos^2 x-\cos x=0$
$\implies \cos x(2\cos x-1)=0$
Either
$\cos x=0$
$\implies x=\dfrac{\pi}{2}\pm 2n\pi$
Or
$2\cos x-1=0$
$\implies \cos x=\dfrac{1}{2}$
$\implies x=\dfrac{\pi}{3}\pm 2n\pi$
where
 $n \in Z$
A: To make it easier to visualise and solve, you can substitute $u=\cos{x}$.
$$2\cos^2{x}-\cos{x}=0$$
$$2u^2-u=0$$
From here, you can either use the quadratic formula or factorise. I chose to factorise:
$$u(2u-1)=0$$
Hence we have the solutions $u=0,\frac{1}{2}$.
Now, we solve the equations resulting from our substitution $u=\cos{x}$:
$$\cos{x}=0 \tag{1}$$
$$\cos{x}=\frac{1}{2} \tag{2}$$
For equation $(1)$, we obtain:`
$$x=\frac{\pi}{2}+k\pi$$
For equation $(2)$, we obtain:
$$x=2k\pi+\frac{\pi}{3}$$
And
$$x=2k\pi-\frac{\pi}{3}$$
Where $k \in \mathbb{Z}$.
Hence, our general solution is:
$$x=\begin{cases} \frac{\pi}{2}+k\pi \\ 2k\pi+\frac{\pi}{3} \\ 2k\pi-\frac{\pi}{3} \end{cases}$$
A: hint: $2\cos^2 x -\cos x = \cos x(2\cos x - 1)=0$
