# How do I solve sec^2x + 2 tan^2x =4?

Sep 20, 2015

Considering only the angles from $0$ to $2 \setminus \pi$, the solutions are $\setminus \frac{\pi}{4}$ and $\frac{5 \setminus \pi}{4}$. Both solutions have a periodicity of $2 \setminus \pi$.

#### Explanation:

Recall the definitions:

• $\sec \left(x\right) = \frac{1}{\setminus} \cos \left(x\right)$
• $\setminus \tan \left(x\right) = \setminus \sin \frac{x}{\setminus} \cos \left(x\right)$

We can thus write ${\sec}^{2} \left(x\right) + 2 \setminus {\tan}^{2} \left(x\right)$ as

$\frac{1}{\setminus} {\cos}^{2} \left(x\right) + 2 \setminus {\sin}^{2} \frac{x}{\setminus} {\cos}^{2} \left(x\right) = \frac{1 + 2 \setminus {\sin}^{2} \left(x\right)}{\setminus {\cos}^{2} \left(x\right)}$.

We want this expression to equal $4$, so we can multiply for $\setminus {\cos}^{2} \left(x\right)$ and get

$\frac{1 + 2 \setminus {\sin}^{2} \left(x\right)}{\setminus {\cos}^{2} \left(x\right)} = 4 \setminus \iff 1 + 2 \setminus {\sin}^{2} \left(x\right) = 4 \setminus {\cos}^{2} \left(x\right)$.

Subtracting $\setminus {\cos}^{2} \left(x\right)$ from both sides, we have

$1 - \setminus {\cos}^{2} \left(x\right) + 2 \setminus {\sin}^{2} \left(x\right) = 3 \setminus {\cos}^{2} \left(x\right)$,

and since $1 - \setminus {\cos}^{2} \left(x\right)$ equals $\setminus {\sin}^{2} \left(x\right)$, the expression becomes

$3 \setminus {\sin}^{2} \left(x\right) = 3 \setminus {\cos}^{2} \left(x\right)$.

Dividing the whole expression by the right member, we have

$\frac{3 \setminus {\sin}^{2} \left(x\right)}{3 \setminus {\cos}^{2} \left(x\right)} = 1$. Canceling the $3$'s out, and recalling again that $\setminus \tan \left(x\right) = \setminus \sin \frac{x}{\setminus} \cos \left(x\right)$, we finally find out that the previous equation is the same as

$\setminus {\tan}^{2} \left(x\right) = 1$, which is verified (considering only the angles between $0$ and $2 \setminus \pi$) when $\setminus \tan \left(x\right) = 1$, i.e. when $x = \setminus \frac{\pi}{4}$, and when $\setminus \tan \left(x\right) = - 1$, i.e. when $x = \frac{5 \setminus \pi}{4}$