If $$\sec^2 x - 3 \sec x + 2 = 0$$, then the value of $$x(0 < x < 90^\circ)$$ is:

$$\sec^2 x - 3 \sec x + 2 = 0$$

$$=$$>  $$\sec^2x-2\sec x-\sec x+2=0$$

$$=$$>  $$\sec x\left(\sec x-2\right)-1\left(\sec x-2\right)=0$$

$$=$$>  $$\left(\sec x-2\right)\left(\sec x-1\right)=0$$

$$=$$>  $$\sec x-2=0$$  or  $$\sec x-1=0$$

$$=$$>   $$\sec x=2$$   or   $$\sec x=1$$

$$=$$>   $$\cos x=\frac{1}{2}$$   or   $$\cos x=1$$

$$=$$>   $$\cos x=\cos60^{\circ\ }$$   or   $$\cos x=\cos90^{\circ\ }$$

$$=$$>   $$x=60^{\circ\ }$$   or   $$x=90^{\circ\ }$$

Since  $$(0 < x < 90^\circ)$$

$$\therefore\ $$ $$x=60^{\circ\ }$$

Hence, the correct answer is Option C