Подробное решение , пожалуйста

Решите задачу:

$\star \; \; \; \frac{sin+2cosxx}{2sinx+cosx} =\frac{5}{4}\\\\ \frac{sinx+2cosx}{2sinx+cosx} =\left [\frac{:cosx}{:cosx}\right] =\frac{tgx+2}{2tgx+1}=\frac{5}{4}\quad \Rightarrow \; \; 4(tgx+2)=5(2tgx+1) \\\\4tgx+8=10tgx+5\quad \Rightarrow \quad 6tgx=3\; ,\; \; \; \underline {tgx=\frac{1}{2}}\\\\\\\star \; \; \; \frac{2sin^2x+sinx\cdot cosx+3}{cos^2x+2sinx\cdot cosx+1} = \frac{2sin^2x+sinx\cdot cosx+3(sin^2x+cos^2x)}{cos^2x+2sinx\cdot cosx+(sin^2x+cos^2x)} =$

$= \frac{5sin^2x+sinx\cdot cosx+3cos^2x}{2cos^2x+2sinx\cdot cosx+sin^2x} = \left [ \frac{:cos^2x}{:cos^2x} \right ]=\frac{5tg^2x+tgx+3}{2+2tgx+tg^2x} =\\\\= \frac{5\cdot \frac{1}{4} +\frac{1}{2}+3}{2+2\cdot \frac{1}{2}+\frac{1}{4}} = \frac{\frac{5}{4}+\frac{2}{4}+\frac{12}{4}}{2+1+\frac{1}{4}} = \frac{\frac{19}{4}}{\frac{13}{4}} = \frac{19}{13}$