Please explain meaning of exp function how to use it .E.G in this (y = exp((sin^2x + sin^4 x + sin^6 x +........infinity)*ln2) ) part. ?

So first we can find the zeros of the given quadratic:

#x^2-9x+8=(x-1)(x-8)# so our zeroes are #x=1,8#.

Let's now deal with the value of #y#. This requires two different tricks: log rules and a sum of an infinite geometric series.

First we see that
# y = e^(text(infinite sum) * ln(2)) = 2^(text(infinite sum))#

Then we can use the formula for an infinite sum of a geometric series:
#text(infinite sum) = sin^2(x) + sin^4(x) + ... #
#= (sin^2(x))^1 + (sin^2(x))^2 + ... #
#= (sin^2(x))/(1 - sin^2(x)) = (sin^2(x))/(cos^2(x)) = tan^2(x) #

Therefore, we know that
#2^(tan^2(x)) = 1 or 2^(tan^2(x)) = 8#
i.e.
#tan^2(x) = 0 or tan^2(x) = 3 #

We can easily realize which angles these correspond to using the unit circle: #x in (0, pi/3)#. By the given constraints, #x=0# can't be the answer. Therefore, we can plug in the value and find that
#(sin(x) + cos(x))/(sin(x) - cos(x)) = (sqrt(3)+1)/(sqrt(3)-1) #

This can actually be simplified:
#(sqrt(3)+1)/(sqrt(3)-1) * (sqrt(3)+1)/(sqrt3+1) = (4 + 2sqrt(3))/2 #

Hence, we either get #2 + sqrt(3)#.