Quick questions on General solutions of trigonometric equations: $\sin$, $\cos$ and $\tan$

$\sin$ has period $2\pi$ and the reflection symmetry $\sin(\pi - \theta) = \sin\theta$, which is exactly the two-points-per-line picture above. So

$\cos$ has period $2\pi$ and is even: $\cos(-\theta) = \cos\theta$. On the unit circle, $\cos x$ is the horizontal coordinate, so a vertical line at $x = k$ meets the circle at an angle and its negative. So

$\tan$ has period $\pi$, not $2\pi$, because $\tan(\theta + \pi) = \tan\theta$ (diametrically opposite points on the circle give the same tangent). So there is only one branch:

If the equation involves $\sin(kx + c)$ rather than $\sin x$, treat the whole bracket as one variable. Solve the general solution for that bracket first, then unwind to $x$ at the very end, dividing every term (including the $2n\pi$) by $k$.

After writing the general solution, substitute $n = \ldots, -1, 0, 1, 2, \ldots$ and keep only the $\theta$ that fall in the required interval. A number line or quick mental check of each $n$ avoids both missing a solution and including one just outside. Endpoints matter: $[0, 2\pi]$ includes $2\pi$, but $[0, 2\pi)$ does not.

Endpoints matter: $[0, 2 \pi)$ excludes $2 \pi$ while $[0, 2 \pi]$ includes it. Check the question wording.