Which one of these ways to solve a simple Trig equation is more correct?

Question

So I'm looking at two different textbooks here and in the beginner practice questions there's simple stuff like $\sin(2x)=-1/2$ with a standard domain of $0$ to $3\pi$.

In one of the textbooks the results for X are simply $x=7\pi/12$ and $x=11\pi/12$.

In the other one the domain gets multiplied by $2$, because of $2x$ and is now $0$ to $2x$ to $6\pi$ and the results for X are now $7\pi/12$, $11\pi/12$, $19\pi/12$, $23\pi/12$, $31\pi/12$ and $35\pi/12$.

Which one should I use? First or second one? Thanks in advance.

Answer 1

Since $\sin(2x)$ is a periodic function whose period is $\pi$, if $\frac{7\pi}{12}$ and $\frac{11\pi}{12}$ are solutions, then, since both of them belong to $[0,\pi]$, there must be another two solutions in $[\pi,2\pi]$ and yet another two in $[2\pi,3\pi]$.

Answer 2

$$\sin2x=-\frac12=\sin\left(-\dfrac\pi6\right)$$

$$2x=n\pi+(-1)^n\left(-\dfrac\pi6\right)$$ where $n$ is any integer

If $n$ is even $=2m$(say)

$$2x=2m\pi-\dfrac\pi6=(12m-1)\dfrac\pi6$$

We need $$0\le(12m-1)\dfrac\pi6\le6\pi $$

$$\iff0\le12m-1\le36\implies1\le m\le3$$

What if $n$ is odd $=2m+1$(say)?