If y is some number between -1 and 1, then solving either y = cos(x) or y = sin(x) for x yields a solution of the form
x = cos⁻¹(y) + 2kπ
or
x = sin⁻¹(y) + 2kπ
where k is any integer.
So we first rewrite the given solution set to resemble either of these templates:
x = π/8 + π/2 k or x = -π/8 + π/2 k
4x = π/2 + 2πk or 4x = -π/2 + 2πk
Now, recall that sin(π/2) = 1 and sin(-π/2) = -1, so that
4x = sin⁻¹(1) + 2πk or 4x = sin⁻¹(-1) + 2πk
Then taking the sine of both sides gives
sin(4x) = 1 or sin(4x) = -1
sin(4x) - 1 = 0 or sin(4x) + 1 = 0
Recall that ab = 0 when either a = 0 or b = 0. Then we can combine both equations into one as
(sin(4x) - 1) (sin(4x) + 1) = 0
and simplifying the right side, we get
sin²(4x) - 1 = 0
We can simplify the equation further by using the Pythagorean identity,
cos²(x) + sin²(x) = 1
so that the equation we ended up with turns into
-cos²(4x) = 0
or simply
cos²(4x) = 0