Consider inequality $$\sin(3\theta)-\frac{\sqrt{3}}{2}\geq0$$ We can proceed to solve it as $$\sin(3\theta)\geq\frac{\sqrt{3}}{2}$$ For simplicity, let's say that $\Theta=3\theta$. $$\sin(\Theta)\geq\frac{\sqrt{3}}{2}$$ By inspecting the unit circle we conclude that $\Theta \in \left\{\frac{\pi}{3},\frac{2\pi}{3}\right\}$ for $[0,2\pi)$ if it were that $\sin(\Theta)=\frac{\sqrt{2}}{3}$. But since we are interested in $\sin(\Theta)\geq\frac{\sqrt{3}}{2}$, we must look for when $\sin(\Theta)$ increases from points $\frac{\pi}{3}$ and $\frac{2\pi}{3}$. By doing that we conclude that for $\sin(\Theta)\geq\frac{\sqrt{3}}{2}$ and the interval $[0,2\pi)$ it is $\Theta \in \left[\frac{\pi}{3},\frac{2\pi}{3}\right]$.
Now, we can generalize this interval even further. For any interval it will be: $$\Theta\in\left[\frac{\pi}{3}+2k\pi,\frac{2\pi}{3}+2k\pi\right]$$ for $k\in\Bbb{Z}$.
To get the interval for $\theta$, we simply multiply everything by $\frac{\theta}{\Theta}=\frac{1}{3}$. $$\theta\in\left[\frac{\pi}{9}+\frac{2k\pi}{3},\frac{2\pi}{9}+\frac{2k\pi}{3}\right]$$ for $k\in\Bbb{Z}$. This is our solution.
What I don't understand is why does $k$ gets multiplies with $\frac{\theta}{\Theta}$. I understand it for angles: since $\Theta=3\theta$, every $\Theta$ angle is $3\cdot\theta$, and therefore every $\theta$ angle is $\Theta/3$.
Why the following isn't the correct solution? $$\theta\in\left[\frac{\pi}{9}+2k\pi,\frac{2\pi}{9}+2k\pi\right]$$ I hope you understand: why does the "rate" or "density" of angles that are valid changes too?
Let $\Theta\in\left[\frac{\pi}{3}+2k\pi,\frac{2\pi}{3}+2k\pi\right]$. This means that $$\Theta = \alpha + 2k\pi$$ for some $\alpha\in[\frac\pi3,\frac{2\pi}3]$ and some $k\in\Bbb Z$. Now, since $\Theta = 3\theta$, we have $$3\theta = \alpha + 2k\pi,$$ so $$\theta = \frac13(\alpha + 2k\pi).$$