Ok, let's assume it's "sec^2 x"
[tex] 8\sec^2x-6\sec x+1=0\\
8\sec^2x-2\sec x-4\sec x+1=0\\
2\sec x(4\sec x-1)-1(4\sec x-1)=0\\
(2\sec x-1)(4\sec x-1)=0\\
2\sec x-1=0\\
2\sec x=1\\
\sec x=\dfrac{1}{2}\\
x\in\emptyset\\\\
4\sec -1=0\\
4\sec =1\\
\sec x =\dfrac{1}{4}\\
x\in\emptyset
[/tex]
So, the number of solutions (real ones) is 0.
