# How do you solve secx/cosx - 1/2secx= 0?

Aug 20, 2015

There are no real solutions

#### Explanation:

$\sec A = \frac{1}{\cos} A$

$\sec \frac{x}{\cos} x - \frac{1}{2} \sec x = 0$
$\sec x \left(\sec x - \frac{1}{2}\right) = 0$
$\sec x = 0$ or $\sec x = \frac{1}{2}$

The range of $| \sec x | \ge 1$, so there are no real solutions

graph{sec x [-10, 10, -5, 5]}

Aug 26, 2015

$x = \arccos \left(2\right)$

$x$ is approximately $1.317 i$

NOTE: This is not a real solution as the other poster dani83 states. If you are restricting yourself to the real numbers then dani83 is correct.

#### Explanation:

Rewrite
NOTE: $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

$\frac{1}{\cos} x \left(\frac{1}{\cos} x\right) - \frac{1}{2} \sec \left(x\right) = 0$

This is just

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

Factor

$\sec x \left(\sec x - \frac{1}{2}\right) = 0$

$\sec x$ cannot be zero so

$\sec x - \frac{1}{2} = 0$

$\sec x = \frac{1}{2}$

Take reciprocal

$\cos x = 2$

$\arccos \left(\cos x\right) = \arccos \left(2\right)$

$x = \arccos \left(2\right)$

This is not a real solution as the other poster says