Sinx-2cosx=2 решите даю 40 баллов

Решите задачу:

$sinx-2cosx=2\; |:\sqrt5\\\\\frac{1}{\sqrt5}\, sinx-\frac{2}{\sqrt5}\, cosx=\frac{2}{\sqrt5}\\\\Tak\; kak\; \; (\frac{1}{\sqrt5})^2+(\frac{2}{\sqrt5})^2=\frac{5}{5}=1\; ,\; to\; \; \frac{1}{\sqrt5}=cosa\; ,\; \frac{2}{\sqrt5}=sina\; \Rightarrow \\\\tga=\frac{sina}{cosa}=2\; \; \to \; \; a=arctg2\\\\cosa\cdot sinx-sina\cdot cosx=\frac{2}{\sqrt5}\\\\sin(x-a)=\frac{2}{\sqrt5}\\\\x-a=(-1)^{n}\cdot arcsin\frac{2}{\sqrt5}+\pi n\; ,\; n\in Z\\\\x=(-1)^{n}\cdot arcsin\frac{2}{\sqrt5}+a+\pi n\; ,\; n\in Z$

$x=(-1)^{n}\cdot arcsin\frac{2}{\sqrt5}+arctg2+\pi n\; ,\; n\in Z$