finding maximum value of triangle length

Question

After learning the applications of derivatives in finding maximum and minimum values I tried to conjure my own example. Now, say we have a triangle with two legs of equal length; what value must the angle included between the two equal-lengthed legs be in order to produce the greatest opposite side? I first wrote down the opposite length as a function of the angle (call the length opposite our variable angle $c$, call the angle itself $x$ and call the legs of equal length $a$): $c(x)^2 = 2a^2 - 2a^2\cos x$. This is the relation known as the law of cosines. Since the magnitude of the angle repeats periodically every $360$ degrees, we can say that it varies between $0$ and $360$ degrees. Intuitively thinking I expect to get the angle at which the maximum length of $c$ is produced to be $180$ degrees. So then, as a function of the angle, $c(x) = (2a^2 - 2a^2\cos x)^{1\over2}$. This maximum value of $c$ should be found at either $x = 0$, or $x = 360$, or at which $c\prime(x)$ is equal to zero. Since I don't expect it to be at $x = 0$ or $x = 360$, I take the derivative of $c(x)$ which is $= $$1\over2$$(2a^2 - 2a^2\cos x)^{-1\over2}2a^2\sin x$. Equating this to zero, I aquire $\sin x = 0$. And sine is $0$ at exactly two angles within a $0$ to $360$ degree range. These angles are $0$ and $180$, hence by trial we find that $0$ is a minimum value, and $180$ is a maximum. I only need confirmation if my reasoning was correct. If not, please explain why. Thank you in advance.