# How do I solve for 0 <= x < 2pi? tan x - sec x +1 = 0

May 19, 2018

x = 0

#### Explanation:

$\tan x - \sec x + 1 = 0$
$\sin \frac{x}{\cos x} - \frac{1}{\cos x} + 1 = 0$
$\sin x - 1 + \cos x = 0$ (1) (condition $\cos x \ne 0 , \mathmr{and} x \ne \frac{\pi}{2}$)
Reminder:
$\sin x + \cos x = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right)$
Equation (1) becomes:
$\sin x + \cos x = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right) = 1$
$\cos \left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$
Trig unit circle and property of cos x function -->
$x - \frac{\pi}{4} = \pm \frac{\pi}{4}$
a. $x - \frac{\pi}{4} = \frac{\pi}{4}$
$x = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$
This answer is rejected because of the above condition: $x \ne \frac{\pi}{2}$
b. $x - \frac{\pi}{4} = - \frac{\pi}{4}$
x = 0.
Check.
x = 0 --> tan x = 0 --> cos x = 1 --> sec x = 1
tan x - sec x + 1 = 0 - 1 + 1 = 0. Proved.