Solving g'(x) = 2 for tan(x) on an interval

The derivative of $\tan(x)$ is $g'(x) = \sec^2(x)$, one of the six memorized trigonometric derivative rules. Setting this equal to 2 gives the equation $\sec^2(x) = 2$ that must be solved for $x$ in $[0, 2\pi)$.

Since $\sec^2(x) = 2$, we have $\cos^2(x) = 1/2$. Using the identity $\sec^2(x) = 1 + \tan^2(x)$, we get $\tan^2(x) = 1$, so $\tan(x) = \pm 1$. Since $\sec^2(x)$ depends only on $\cos^2(x)$, the equation $\sec^2(x) = 2$ is satisfied whenever $\cos(x) = \pm 1/\sqrt{2}$, giving four candidate angles: $x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$.

We can verify these are related by periodicity: when $\tan(x) = 1$, the principal solution in $[0, 2\pi)$ is $x = \arctan(1) = \pi/4$, and the next solution (shifted by the period $\pi$) is $x = \pi/4 + \pi = 5\pi/4$. When $\tan(x) = -1$, the principal solution is $x = \pi + \arctan(-1) = 3\pi/4$, and the next solution is $x = 3\pi/4 + \pi = 7\pi/4$.

Thus, on $[0, 2\pi)$, the four solutions where $g'(x) = 2$ are $x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$. In terms of $\arctan$, these can be expressed as $x = \arctan(1)$ and $x = \arctan(1) + \pi$ (the two solutions where $\tan(x) = 1$), and $x = \pi + \arctan(-1)$ and $x = \pi + \arctan(-1) + \pi$ (the two solutions where $\tan(x) = -1$).