$\begin{gathered}

  \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {\sin x} \right]\left( {\sin x - 1} \right)}}{{\sin x + cos2x}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {\sin x} \right]\left( {\sin x - 1} \right)}}{{\sin x + 1 - 2{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {\sin x} \right]\left( {\cancel{{\sin x - 1}}} \right)}}{{\left( { - 2\sin x - 1} \right)\left( {\cancel{{\sin x - 1}}} \right)}} \hfill \\

  \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {\sin x} \right]}}{{ - 2\sin x - 1}}\left\{ {\begin{array}{*{20}{c}}

  {\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \frac{{\left[ {\sin x} \right]}}{{ - 2\sin x - 1}} = \frac{{\left[ {{1^ - }} \right]}}{{ - 3}} = 0} \\ 

  {\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ - }} \frac{{\left[ {\sin x} \right]}}{{ - 2\sin x - 1}}\frac{{\left[ {{1^ - }} \right]}}{{ - 3}} = 0} 

\end{array}} \right. \hfill \\ 

\end{gathered} $