I understand why $\sin(x)=\frac{\sqrt{3}}{2}$ has two answers, $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ but I don't understand why $\tan(x)=1$ only has one solution (according to my book and other places I've looked online) $\frac{\pi}{4}$. Why isn't $\frac{5\pi}{4}$ also a solution given that tan is positive in Q1 and Q3 and at $\frac{5\pi}{4}$ both $\sin$ and $\cos$ are $-\frac{\sqrt{2}}{2}$
Note that the interval is not given.
If the interval is not given, you would be wise to assume it's from minus infinity to positive infinity.
Consequently you should give the general solution.
So, $$\sin(x)=\frac{\sqrt{3}}{2}$$ has solutions, $$\frac{\pi}{3}+2\pi k$$ or $$\frac{2\pi}{3}+2\pi k$$ for integer values of $k$
Shall I leave you to do similar for the other equations in your question ?
Top Tip : Drawing a graph of the appropriate trigonometric function is helpful.