$\begin{array}{l} t = 4\sin x + 3\cos x = 5\left( {\dfrac{4}{5}\sin x + \dfrac{3}{5}\cos x} \right)\\ = 5\sin \left( {x + \alpha } \right)\left( {\alpha = \arccos \left( {\dfrac{4}{5}} \right)} \right)\\ \Rightarrow - 5 \le t \le 5\\ PT \Leftrightarrow t + \dfrac{6}{{t + 1}} = 6\\ \Leftrightarrow {t^2} + t + 6 = 6t + 6\\ \Leftrightarrow {t^2} - 5t = 0 \Leftrightarrow t\left( {t - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = 0\\ t = 5 \end{array} \right.(TM)\\ \Leftrightarrow \left[ \begin{array}{l} 5\sin \left( {x + \alpha } \right) = 0\\ 5\sin \left( {x + \alpha } \right) = 5 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \sin \left( {x + \alpha } \right) = 0\\ \sin \left( {x + \alpha } \right) = 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x + \alpha = k\pi \\ x + \alpha = \dfrac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - \alpha + k\pi \\ x = \dfrac{\pi }{2} - \alpha + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$
