\({\cos ^2}\left( {x + 1} \right)\lg \left( {9-2x-{x^2}} \right) \ge 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\cos ^2}\left( {x + 1} \right) \cdot \lg \left( {10-{{\left( {x + 1} \right)}^2}} \right) \ge 1.\)
Так как \(0 \le {\cos ^2}\left( {x + 1} \right) \le 1\) и \(\lg \left( {10-{{\left( {x + 1} \right)}^2}} \right) \le 1\), то неравенство выполнится только в случае:
\(\left\{ {\begin{array}{*{20}{c}}{{{\cos }^2}\left( {x + 1} \right) = 1,\,\,\,\,\,\,\,\,\,\,\,}\\{\lg \left( {10-{{\left( {x + 1} \right)}^2}} \right) = 1}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\sin \left( {x + 1} \right) = 0,\\x = -1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -1.\)
Ответ: \(-1.\)