\(\left( {3-{{\cos }^2}x-2\sin x} \right)\left( {{{\lg }^2}y + 2\lg y + 4} \right) \le 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {3-1 + {{\sin }^2}x-2\sin x} \right)\left( {{{\left( {\lg y + 1} \right)}^2} + 3} \right) \le 3\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left( {{{\left( {\sin x-1} \right)}^2} + 1} \right)\left( {{{\left( {\lg y + 1} \right)}^2} + 3} \right) \le 3.\)
Так как \({\left( {\sin x-1} \right)^2} + 1 \ge 1\) и \({\left( {\lg y + 1} \right)^2} + 3 \ge 3\), то неравенство выполнится только в случае:
\(\left\{ {\begin{array}{*{20}{c}}{{{\left( {\sin x-1} \right)}^2} + 1 = 1,}\\{{{\left( {\lg y + 1} \right)}^2} + 3 = 3\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = \dfrac{{\rm{\pi }}}{2} + 2{\rm{\pi }}n,\\y = \dfrac{1}{{10}}\end{array} \right.\,\,\,\,n \in Z.\)
Ответ: \(\left( {\,\dfrac{{\rm{\pi }}}{2} + 2{\rm{\pi }}\,n;\,\dfrac{1}{{10}}} \right),\;\;n \in Z.\)