The minimum value of
$\frac{7}{4\sin x+3\cos x-6} is$

$\mathrm{acos} \mathrm{x}+\mathrm{bsin} \mathrm{x}+\mathrm{c}$ minimum, maximum values are  $\mathrm{c}-\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$,  $\mathrm{c}+\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

minimum value of  $\frac{7}{4\sin x+3\cos x-6}$

$\begin{array}{rr}=\frac{7}{m (4\sin x+3\cos x-6)}& \end{array}$

$\begin{array}{rr}=\frac{7}{-6+\sqrt{16+9}}& \end{array}$

$=\frac{7}{-6+5}=-7$