Write a trigonometric equation whose solution is the set given: {x|x=(pi/8)+(pi/2k) or x=-pi/8+pi/2k}, k ∈ Z

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