We're given that:

f(x) = sin x + cos x

We need to find the value of x such that:

f(x) + f(-1) = 0

Substituting f(x) and simplifying, we get:

sin x + cos x + sin(-1) + cos(-1) = 0

Using the identity sin(-x) = -sin(x) and cos(-x) = cos(x), we get:

sin x + cos x - sin 1 - cos 1 = 0

Rearranging, we get:

sin x + cos x = sin 1 + cos 1

Now, we can substitute f(x) back into this equation and simplify:

f(x) = sin x + cos x = sin 1 + cos 1

Therefore, we have:

sin x + cos x = sin 1 + cos 1

sin x = sin 1 and cos x = cos 1

Since sin x = sin 1, we have:

x = nπ + (-1)ⁿ * 1, where n is any integer.

Similarly, since cos x = cos 1, we have:

x = 2nπ ± 1, where n is any integer.

Therefore, the possible solutions for x are:

x = nπ + (-1)ⁿ * 1 or x = 2nπ ± 1, where n is any integer.