# What are the solutions to 10(sec^2)x+5(tan^2)x-15=0 over the interval (0,2pi)?

May 11, 2018

pi/6; (2pi)/3; (7pi)/6; (5pi)/3

#### Explanation:

$10 {\sec}^{2} x + 5 {\tan}^{2} x - 15 = 0$
Replace ${\sec}^{2} x$ by $\left(1 + {\tan}^{2} x\right)$ from trig identity.
$10 \left(1 + {\tan}^{2} x\right) + 5 {\tan}^{2} x - 15 = 0$
$10 {\tan}^{2} x + 5 {\tan}^{2} x - 5 = 0$
$15 {\tan}^{2} x = 5$ --> ${\tan}^{2} x = \frac{1}{3}$
$\tan x = \pm \frac{\sqrt{3}}{3}$
Trig table and unit circle give 4 solutions for tan x -->
a. $\tan x = \frac{\sqrt{3}}{3}$ -->
$x = \frac{\pi}{6}$ and $x = \frac{\pi}{6} + \pi = \frac{7 \pi}{6}$
b. $\tan x = - \frac{\sqrt{3}}{3}$ -->
$x = \frac{2 \pi}{3}$ and $x = \pi + \frac{2 \pi}{3} = \frac{5 \pi}{3}$