# Question #7d61a

May 13, 2017

$x = \frac{\pi}{4} + \frac{n \pi}{2}$ where $n \in \mathbb{Z}$

#### Explanation:

Question: ${\sec}^{2} x + {\tan}^{2} x = 3 {\tan}^{2} x$

We can move the ${\tan}^{2} x$ to the right-hand side:
${\sec}^{2} x = 2 {\tan}^{2} x$

Note that $\sec x = \frac{1}{\cos} x$ and $\tan x = \sin \frac{x}{\cos} x$
So we have:
$\frac{1}{\cos} ^ 2 x = \frac{2 {\sin}^{2} x}{\cos} ^ 2 x$
$\frac{1}{\cos} ^ 2 x - \frac{2 {\sin}^{2} x}{\cos} ^ 2 x = 0$
$\frac{1 - 2 {\sin}^{2} x}{\cos} ^ 2 x = 0$

Note the double angle identity for $\cos \left(2 x\right) = {\cos}^{2} x - {\sin}^{2} x = 1 - 2 {\sin}^{2} x$
So we have:
$\frac{{\cos}^{2} x - {\sin}^{2} x}{\cos} ^ 2 x = 0$

${\cos}^{2} \frac{x}{\cos} ^ 2 x - {\sin}^{2} \frac{x}{\cos} ^ 2 x = 0$

$1 - {\sin}^{2} \frac{x}{\cos} ^ 2 x = 0$

$1 - {\tan}^{2} x = 0$

${\tan}^{2} x = 1$

$\tan x = \pm \sqrt{1}$

$\tan x = \pm 1$

Therefore,
$x = \frac{\pi}{4} + \frac{n \pi}{2}$ where $n \in \mathbb{Z}$