If f(x)=sinx and g(x)=cosx, then the set for all x for which f'(x)=g'(x) is:

A. Pi/4 + k(pi)
B. Pi/2 + k(pi)
C. 3pi/4 +k(pi)
D. Pi/2 + 2k(pi)
E. 3pi/2 + 2k(pi)

f(x) = sinx ---> f ' (x) = cosx

g(x) = cosx ---> g ' (x) = -sinx

so we want -sinx = cosx
-sinx/cosx = 1
tanx = -1
I know tan 3π/4 = -1 and tan 7π/4 = -1
and the period of tanx is π

since 3π/4 + π = 7π/4 we can just state that first answer, and

x = 3π/4 + kπ , where k is an integer

To find the set of values for x that satisfy f'(x) = g'(x), we need to find the derivative of the functions f(x) = sin(x) and g(x) = cos(x).

Let's start by finding the derivative of f(x) = sin(x). The derivative of sin(x) with respect to x is cos(x), so f'(x) = cos(x).

Next, let's find the derivative of g(x) = cos(x). The derivative of cos(x) with respect to x is -sin(x), so g'(x) = -sin(x).

Now, we can set f'(x) equal to g'(x) and solve for x:
cos(x) = -sin(x)

Dividing both sides by cos(x), we get:
tan(x) = -1

The equation tan(x) = -1 has solutions at specific values of x, specifically at x = -π/4 + kπ, where k is an integer.

So, the set of values for x that satisfy f'(x) = g'(x) is x = -π/4 + kπ.

Comparing the answer choices given:
A. π/4 + kπ: This option does not match the solution.
B. π/2 + kπ: This option does not match the solution.
C. 3π/4 + kπ: This option does not match the solution.
D. π/2 + 2kπ: This option matches the solution.
E. 3π/2 + 2kπ: This option does not match the solution.

Therefore, the correct answer is D. π/2 + 2kπ.