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

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

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Answer 1 · 2018-05-19T21:32:16

x = 0

Explanation:

$tan x - sec x + 1 = 0$ $tan x - sec x + 1 = 0$
$\frac{sin x}{cos x} - \frac{1}{cos x} + 1 = 0$ $sin x/(cos x) - 1/(cos x) + 1 = 0$
$sin x - 1 + cos x = 0$ $sin x - 1 + cos x = 0$ (1) (condition $cos x \neq 0, and x \neq \frac{π}{2}$ $cos x!= 0,and x != pi/2$ )
Reminder:
$sin x + cos x = \sqrt{2} cos (x - \frac{π}{4})$ $sin x + cos x = sqrt2cos (x - pi/4)$
Equation (1) becomes:
$sin x + cos x = \sqrt{2} cos (x - \frac{π}{4}) = 1$ $sin x + cos x = sqrt2cos (x - pi/4) = 1$
$cos (x - \frac{π}{4}) = \frac{1}{\sqrt{2}}$ $cos (x - pi/4) = 1/sqrt2$
Trig unit circle and property of cos x function -->
$x - \frac{π}{4} = \pm \frac{π}{4}$ $x - pi/4 = +- pi/4$
a. $x - \frac{π}{4} = \frac{π}{4}$ $x - pi/4 = pi/4$
$x = \frac{π}{4} + \frac{π}{4} = \frac{π}{2}$ $x = pi/4 + pi/4 = pi/2$
This answer is rejected because of the above condition: $x \neq \frac{π}{2}$ $x != pi/2$
b. $x - \frac{π}{4} = - \frac{π}{4}$ $x - pi/4 = - pi/4$
x = 0.
Check.
x = 0 --> tan x = 0 --> cos x = 1 --> sec x = 1
tan x - sec x + 1 = 0 - 1 + 1 = 0. Proved.

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

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