Question

No. Of roots in eq 8sec^x-6secx+1=0

Answer 1

Ok, let's assume it's "sec^2 x"

$8\sec^2x-6\sec x+1=0\\ 8\sec^2x-2\sec x-4\sec x+1=0\\ 2\sec x(4\sec x-1)-1(4\sec x-1)=0\\ (2\sec x-1)(4\sec x-1)=0\\ 2\sec x-1=0\\ 2\sec x=1\\ \sec x=\dfrac{1}{2}\\ x\in\emptyset\\\\ 4\sec -1=0\\ 4\sec =1\\ \sec x =\dfrac{1}{4}\\ x\in\emptyset$

So, the number of solutions (real ones) is 0.