General solution of $\sin 2x+\cos x=0$ 
Solve the trigonometric equation $\sin 2x+\cos x=0$

My Attempt
$$
2\sin x\cos x+\cos x=0\implies\cos x=0 \text{ or }\sin x=\frac{-1}{2}=\sin\frac{-\pi}{6}\\
x=(2n+1)\frac{\pi}{2} \text{ or }x=n\pi+(-1)^n\frac{-\pi}{6}\\
x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}
$$
Reference
$$
\cos x=-\sin 2x=\cos\Big(\frac{\pi}{2}+2x\Big)\implies x=2n\pi\pm\Big(\frac{\pi}{2}+2x\Big)\\
-x=2n\pi+\frac{\pi}{2}\text{ or }3x=2n\pi-\frac{\pi}{2}\implies x=2m\pi-\frac{\pi}{2}\text{ or }x=\frac{2m\pi}{3}-\frac{\pi}{6}
$$
But my reference gives the solution $x=2n\pi-\dfrac{\pi}{2}$ or $x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}$. I understand how it is reached and both represent the same solutions. But, how do I derive the solution in my reference from what I found in my attempt ? i.e.,

How to derive
  $$
\bigg[x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}\bigg]\\
\implies \bigg[x=2n\pi-\dfrac{\pi}{2}\text{ or }x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}\bigg]
$$

 A: There are two ways to solve the equation.
Your method: $2\sin x\cos x+\cos x=0$, so $\cos x(2\sin x+1)=0$. Thus we have either $\cos x=0$ or $\sin x=-1/2$. Thus 
\begin{align}
x&=\frac{\pi}{2}+2n\pi &\text{or}&&  x&=-\frac{\pi}{2}+2n\pi && \text{(from $\cos x=0$)} \\[4px]
x&=-\frac{\pi}{6}+2n\pi  &\text{or}&& x&=\frac{7\pi}{6}+2n\pi && \text{(from $\sin x=-1/2$)}
\end{align}
(you grouped together the families of solutions of $\cos x$ and of $\sin x=-1/2$, but I prefer to keep them separated). Good job.
Alternative method: $\cos x=-\sin2x=\cos(\frac{\pi}{2}+2x)$. Therefore either
$$
x=\frac{\pi}{2}+2x+2n\pi \to x=-\frac{\pi}{2}-2n\pi
$$
or
$$
x=-\frac{\pi}{2}-2x+2n\pi \to 3x=-\frac{\pi}{2}+2n\pi \to x=\frac{2n\pi}{3}-\frac{\pi}{6}
$$
How do you recover the previous sets of solutions?
One set is already present. For the other three, consider the cases when $n=3k$, $n=3k+1$ or $n=3k+2$, with integer $k$. Then
\begin{align}
n&=3k & x&=\frac{6k\pi}{3}-\frac{\pi}{6}=2k\pi-\frac{\pi}{6} \\[4px]
n&=3k+1 & x&=\frac{(6k+2)\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{2\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{\pi}{2}\\[4px]
n&=3k+2 & x&=\frac{(6k+4)\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{4\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{7\pi}{6}
\end{align}
A: We have that


*

*$\cos x = 0 \implies x=\frac{\pi}2+k\pi$

*$\sin x = -\frac12 \implies x=-\frac{\pi}6+2k\pi \quad \lor \quad x=-\frac{5\pi}6+2k\pi$
which is equivalent to


*

*$x=-\frac{\pi}2+2k\pi$

*$x=-\frac{\pi}6+\frac23k\pi$
to see that draw the solution points on the trigonometric circle.

A: You and your reference (assuming no typos) have omitted many solutions.  For instance, $x = \pi/2$, which you are missing, and $-\pi/2$, which your reference is missing.
Assuming throughout that $k$ is any integer.
From $\cos(x) = 0$, $x = \pm \cos^{-1}(0) + 2\pi k$, giving the two solution families $x = \pm \frac{\pi}{2} + 2\pi k$.  These can be written as a single family (note the coefficient of $k$): $\frac{\pi}{2} + \pi k$, which is equivalent to $\frac{-\pi}{2} + \pi k$.  This equivalence is most likely what is going on between your solution and your reference's.
From $\sin(x) = \frac{-1}{2}$, $x = \sin^{-1}\left(\frac{-1}{2}\right) + 2\pi k = \frac{-\pi}{6} + 2 \pi k$ or $x = \pi - \sin^{-1}\left(\frac{-1}{2}\right) + 2\pi k = \frac{7 \pi}{6} + 2 \pi k$.  Taken together, you should recognize these as your solution and one of them you quote from your reference.
Are you sure you have copied your reference's entire answer and correctly copied its $k$ coefficients?
A: $$\sin(2x)+\cos(x)=0$$
$$2\sin(x)\cos(x)+\cos(x)=0$$
$$\cos(x)\left(2\sin(x)+1\right)=0$$
so you have two sets of solutions:
$$\cos(x)=0,\,\sin(x)=-\frac{1}{2}$$
EDIT:
firstly they have:
$$x=2n\pi-\pi/2=\pi(2n-1/2)$$
and you have:
$$x=(2n+1)\pi/2$$
let:
$$\pi/2(4n-1)=\pi/2(2m+1)$$
so:
$$4n-1=2m+1$$
$$m=(4n-2)/2=2n-1$$
so for all integer values of $n$, $m$ is also an integer and so they are equivalent?
