Verified answer

Well in general the fixed points will be x(t) = constant, and thus x' = x'' = 0.

Substituting into the original equation, we have:

x'' + 14 x' + 98 cos x = 49

98 cos x = 49

cos x = 1/2.

What are the solutions to this? Well, that is tricky, because the function f = cos θ is periodic and doesn't have a complete inverse. Instead, there is an inverse valid for 0 ≤ θ ≤ π, which we call the "arccosine" or "acos" function. But that is not the only solution, because of these two symmetries:

rule 1: cos x = cos -x

rule 2: cos x = cos(x + 2π)

Thus, we know that the only solution between 0 ≤ x ≤ π is x = acos(1/2). It turns out that this number is π/3, because cos(π/3) = cos(60°) = 1/2. [draw out the equilateral triangle to see this.]

We use rule 1 on this result: So, we know that there are two solutions between -π ≤ x ≤ π, and they are -π/3 and +π/3.

Now we use rule 2 to translate this solution to the entire real line. We can add or subtract any discrete multiple of 2π, and thus we have:

x = 2 π k ± π / 3, k ∈ Z.

Notice that the acceptable k's are not real numbers, but are integers. +4π and -100π are allowed, but 22.68π is not.