Trigonometry has some complex stuff, but sometimes students have trouble with the beginning, the basics.
As an example, here's what seems to be a simple question:
$$2cos(x)=1$$ $$0°≤x≤360°$$ Find the possible values of $x$
For students, this question may not seem as simple to them. Is there a good way I can explain this question, and these classes of questions in general?
If you can, provide an example question and an example explanation.
It seems like a problem of unclear or inapplicable definitions. As stated in the comments, it sounds like you've given the students the following definition: Given an angle $t$, we consider a right triangle with an angle of measure $t$, and define $\cos(t)$ to be the ratio of the side length of this triangle adjacent to the given angle to the length of the hypotenuse.
The problem is, this definition only makes sense when $0 < t < \pi/2$ (or, if you prefer degrees, $0^\circ < t < 90^\circ$. Indeed, a right triangle's other two angles have measure strictly between $0^\circ$ and $90^\circ$. But you've given the students a problem that assumes cosine has been defined for any $t$ between $0^\circ$ and $360^\circ$, which makes no sense with the definition they've been given!
I would recommend instead defining cosine as the $x$-coordinate on the unit circle. A simple diagram (drawing an appropriate right triangle inside the unit circle) shows this agrees with the other definition for angles in the first quadrant, but the unit circle definition also clearly makes sense for any angle. Plus, it makes it much easier to reason about this kind of question — just draw the line $x = 1/2$ and see where it intersects the unit circle, then use geometric reasoning to find the angles.
I'm still a bit confused about what you mean by "algebraic methods", by the way. Both the triangle definition and the unit circle definition are geometric, so you really can't compute values of cosine using these definitions without appealing to geometry.
For the easy memorizing of special points:
$$\cos(0^\circ)=\frac{\sqrt4}2\\\cos(30^\circ)=\frac{\sqrt3}2\\\cos(45^\circ)=\frac{\sqrt2}2\\\cos(60^\circ)=\frac{\sqrt1}2\\\cos(90^\circ)=\frac{\sqrt0}2$$
Same thing for $\sin(\theta)$ but going backwards. Then it should be easy to see that $x=60^\circ$ is one answer. From there, I would try to show them the unit circle and how it is symmetric. Using that, it should become obvious that $300^\circ$ is the only other answer.
Beyond that, either you are teaching some pretty advanced trigonometry, it happens to be a special special point, like $\cos(15^\circ)$ using the half-angle theorem, or you give them calculators. I can't imagine it being any other way, unless you were teaching a numerical analysis type of class.
If students are struggling to understand a relatively simple trig equation like the one given, I find that graphs often help. We can sketch y=2cos(x) and y=1, and get a feeling for how many solutions to expect in the given interval. Moreover symmetries of the graphs can help determine the full set of solutions, much as the unit circle approach does. Interestingly, I believe in the UK the unit circle development of trig functions isn't such a common approach but many teachers use the (IMO awful) 'CAST' diagram for solving trig equations, devoid of deeper understanding.
