Let $x\in (0,\frac{\pi }{2})$ and ${\log }_{24\sin x}\left(24\cos x\right)=\frac{3}{2},$ then  the value of $\cos e{c}^2\left(x\right)$ is

${\left(24\sin x\right)}^{\frac{3}{2}}=24\cos x\Rightarrow \sqrt{24}{\left(\sin x\right)}^{\frac{3}{2}}=\cos x\Rightarrow 24{\sin }^{3}x={\cos }^{2}x=1-{\sin }^{2}x$

Put $\sin x=t,$ we get $24t^3+t^2-1=0\Rightarrow (3t-1)\underset{>0}{\underbrace{(8t^2+3t+1)}}=0\Rightarrow t=\frac{1}{3}$

$\therefore t=\frac{1}{3}i.e.\sin x=\frac{1}{3}\Rightarrow \cos ecx=3\Rightarrow \cos e{c}^{2}x=9$