trigonometric equations finding $x$, when $\cos(x)=0$ or some value I'm practicing trigonometric equations where we need to find the value of $x$, however sometimes I keep getting a completely different solution than the one in the book and I don't really understand where I am making a mistake (and it's always in finding the $x$ in the end). For example:
Problem: $$\cos x  + \sqrt3\cos 2x + \cos3x=0$$ which gives: $$2\cos2x(\cos x + \frac{\sqrt3}{2})=0$$
and then:  $2\cos2x=0$ , where I get that $x=\frac{\pi}{4}+\frac{k\pi}{2}, k\in \Bbb Z$ and:  $\cos x= -\frac{\sqrt3}{2}$, where I get $x=\frac{5\pi}{6}+ 2k\pi, k\in \Bbb Z$ (which based on my book is the only solution I got right)  
I believe I also have this: $\cos (2\pi - x)= -\frac{\sqrt3}{2} $, and the solution I got: $x=\frac{7\pi}{6}+ 2k\pi, k\in \Bbb Z$
I guess I'm getting something wrong since the solutions given are:  $x=\frac{(2k+1)\pi}{2}$ ; $x=\frac{5\pi}{6}+2k\pi$; $x=-\frac{5\pi}{6}+2k\pi$, $k\in \Bbb Z$
 A: $$\cos(nx)=c$$ has the solutions
$$nx=2k\pi\pm\arccos(c)$$ or
$$x=\frac{2k\pi\pm\arccos(c)}n.$$
In the (frequent) event that $\arccos(c)$ is a rational fraction $\dfrac pq$ of $\pi$,
$$x=\frac{2kq\pm p}{qn}\pi.$$
In particular we have the special cases

*

*$c=-1\to\dfrac11\to \dfrac{2k\pm1}n\pi=\dfrac{2k+1}n\pi$,


*$c=0\to\dfrac12\to \dfrac{4k\pm1}{2n}\pi=\dfrac{2k+1}{2n}\pi$,


*$c=1\to\dfrac01\to \dfrac{2k\pm0}n\pi= \dfrac{2k}n\pi$.
A: hint
If $$\cos(2x)(\cos(x)\frac{\sqrt{3}}{2})=0$$
then
$$\cos(2x)=0=\cos(\frac{\pi}{2})$$
or
$$\cos(x)=-\frac{\sqrt{3}}{2}=\cos(\pi-\frac{\pi}{6})$$
then
$$2x=\pm \frac{\pi}{2}+2k\pi=\frac{\pi}{2}+k\pi$$
or
$$x=\pm \frac{5\pi}{6}+2k\pi$$
The solutions are
$$\frac{\pi}{4}+k\frac{\pi}{2}\;,\;$$
$$ \frac{5\pi}{6}+2k\pi=-\frac{7\pi}{6}+2(k+1)\pi\;$$
and
$$\; \frac{-5\pi}{6}+2k\pi=\frac{7\pi}{6}+2(k-1)\pi$$
A: If $-5\pi/6+2m\pi=7\pi/6+2n\pi$ where $m,n$ are arbitrary integers
$\iff m=n+1$
So, we have actually not missed anything
As $\cos(2r\pi\pm A)=\cos A,$
If $\cos B=\cos C,B=2p\pi\pm C$ where $r,p$ are Integers
