The general solutions of the equation $\tan ^2 \theta+\sec 2 \theta=1$ are

Transforming the equation $\tan^2\theta + \sec 2\theta = 1$ begins with the identity $\tan^2\theta = \sec^2\theta - 1$, yielding:
$\sec^2\theta + \sec 2\theta = 2$

Expressing secants as reciprocals of cosines, $\frac{1}{\cos^2\theta} + \frac{1}{\cos 2\theta} = 2$, and using $\cos 2\theta = 2\cos^2\theta - 1$ to write $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$, substitution gives:
$\frac{2}{1 + \cos 2\theta} + \frac{1}{\cos 2\theta} = 2$

Combining fractions with a common denominator results in $\frac{3\cos 2\theta + 1}{\cos 2\theta (1 + \cos 2\theta)} = 2$. Cross-multiplying and simplifying leads to the quadratic equation:
$2\cos^2 2\theta - \cos 2\theta - 1 = 0$

Factoring as $(2\cos 2\theta + 1)(\cos 2\theta - 1) = 0$ gives solutions $\cos 2\theta = -\frac{1}{2}$ or $\cos 2\theta = 1$.

For $\cos 2\theta = 1$, the general solution is $2\theta = 2n\pi$, so $\theta = n\pi$.

For $\cos 2\theta = -\frac{1}{2}$, the general solution is $2\theta = 2n\pi \pm \frac{2\pi}{3}$, leading to $\theta = n\pi \pm \frac{\pi}{3}$.

Both solution sets are valid since none introduce undefined values for $\tan\theta$ or $\sec 2\theta$.

The general solutions are $\theta = n\pi$ and $\theta = n\pi \pm \frac{\pi}{3}$, for $n \in \mathbb{Z}$.