I've been working on this example in Boyce's Elementary Differential Equations and Boundary Value Problems book. The example shows how to find a solution to $$y''+4y=3\csc(t)$$ through the method of variation of parameters.
We know that the general solution of the homogeneous equation corresponding to the equation above, $y''+4y=0$, is $y_c(t)=c_1\cos(2t)+c_2\sin(2t)$. So the idea of the method is to replace $c_1$ and $c_2$ with the functions $u_1$ and $u_2$, respectively, and then try to determine these functions so that the expression $y=u_1\cos(2t)+u_2\sin(2t)$ is a solution to the non-homogeneous equation. At this point we would calculate the first and second derivatives of this last expression and substitute for $y$ in the non-homogeneous equation, what would lead us to one equation involving a combination of $u_1$ and $u_2$, and their derivatives. That allows us to impose a second condition so that we obtain two equations for two unknown functions $u_1$ and $u_2$.
In the example, they require that $u'_1\cos(2t)+u'_2\sin(2t)=0$. That condition, after we've done the calculations, leads us to the following system: $$\begin{cases} u'_1\cos(2t)+u'_2\sin(2t)=0 \\\\ -2u'_1\sin(2t)+2u'_2\cos(2t)=3\csc(t) \end{cases} $$
Now, the example goes on to solve this system beginning with $$u'_2=-u'_1\frac{\cos(2t)}{\sin(2t)}$$ Since we can't divide by zero, we have to suppose also that $\sin(2t)\ne0$, which implies $t\ne n\frac{\pi}{2}, n \in \mathbb{Z}$. But the book doesn't make this observation. It just moves on to find both $u'_1$ and $u'_2$ and, after integration, declares that $u_1=-3\sin(t)+c_1$ and $u_2=\frac{3}{2}\ln{|\csc(t)-\cot(t)|}+3\cos(t)+c_2$. These two last results, after some more manipulations, give us the general solution to our differential equation: $$y=3\sin(t)+\frac{3}{2}\ln{|\csc(t)-\cot(t)|}\sin(2t)+c_1\cos(2t)+c_2\sin(2t)$$
My question lies in that consideration not made by the book of $\sin(2t)$ not being equal to zero. I think that if we were solving a system of equations problem, not a differential equation one, taking into consideration only the system above, we would have to analyze both the case when $\sin(2t)\ne0$ and the case when $\sin(2t)=0$. Why doesn't the book take this restriction into account? Doesn't that impact at all the interval of validity of the general solution found at the end?
OBS: I'm taking my first course on differential equations, so I may be missing some key information or insight here.