# How do you solve sec^2x-2tan^2x=0?

Oct 25, 2016

$x = \frac{\pi}{4} + \frac{k \pi}{2} , k \in \mathbb{Z}$

#### Explanation:

Here, we can use the Pythagorean Identity, which states that $1 + {\tan}^{2} x = {\sec}^{2} x$.

We can substitute this into the original equation to write it just in terms of tangent, instead of with both tangent and secant.

$\left({\sec}^{2} x\right) - 2 {\tan}^{2} x = 0 \text{ "=>" } \left(1 + {\tan}^{2} x\right) - 2 {\tan}^{2} x = 0$

Simplifying this by combining the ${\tan}^{2} x$ terms gives:

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

Thus:

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

Taking the square root, and remembering both the positive and negative roots:

$\tan x = \pm 1$

Notice that $\tan x = 1$ at $x = \frac{\pi}{4}$ and $x = \frac{5 \pi}{4}$, and all angles with a reference angle of $\frac{\pi}{4}$ in quadrants $\text{I}$ and $\text{III}$.

Similarly, $\tan x = - 1$ at $x = \frac{3 \pi}{4}$ and $\frac{7 \pi}{4}$ and all other angles with a reference angle of $\frac{\pi}{4}$ in quadrants $\text{II}$ and $\text{IV}$.

Combining all these answers, and remembering that they go on forever, we see that $x = \frac{\pi}{4} , \frac{3 \pi}{4} , \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{9 \pi}{4}$, with $x = \frac{9 \pi}{4}$ being exactly one $2 \pi$ revolution away from $\frac{\pi}{4}$.

But, the solutions don't stop, and we can generalize them by making a rule. Note that every solution for $x$ is $\frac{2 \pi}{4} = \frac{\pi}{2}$ away from the next solution. So, we can think about starting at $x = \frac{\pi}{4}$ then adding some whole number amount of $\frac{\pi}{2}$ values to the starting value.

This is expressed as

$x = \frac{\pi}{4} + \frac{k \pi}{2} , k \in \mathbb{Z}$

Note that $k \in \mathbb{Z}$ just means that $k$ is an integer, which includes numbers like $1 , - 5 , 7 ,$ and $- 68$.