令 $\tan \dfrac x2=t$，则$$y=\dfrac{3+5\sin x}{\sqrt{5+4\cos x+3\sin x}}=\dfrac{3+5\cdot \dfrac{2t}{1+t^2}}{\sqrt{5+4\cdot\dfrac{1-t^2}{1+t^2}+3\cdot\dfrac{2t}{1+t^2}}}=\dfrac{t+3}{|t+3|}\cdot \dfrac{3t+1}{\sqrt{1+t^2}},$$考虑到函数 $\varphi(t)=\dfrac{3t+1}{\sqrt{1+t^2}}$ 的导函数$$\varphi'(t)=\dfrac{3-t}{(1+t^2)\sqrt{1+t^2}},$$于是 $\varphi(t)$ 在 $(-3,3)$ 上单调递增，在 $(3,+\infty)$ 上单调递减．又因为当 $t<0$ 时，$$|\varphi(t)|<\dfrac{|3t|}{\sqrt{1+t^2}}<3,$$所以函数 $y=\dfrac{t+3}{|t+3|}\cdot \varphi(t)$ 的图象如图，所求的值域为 $\left(-\dfrac {4\sqrt{10}}5,\sqrt{10}\right]$．