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

1 Answer
Nghi N
May 19, 2018

x = 0

Explanation:

#tan x - sec x + 1 = 0#
#sin x/(cos x) - 1/(cos x) + 1 = 0#
#sin x - 1 + cos x = 0# (1) (condition #cos x!= 0,and x != pi/2#)
Reminder:
#sin x + cos x = sqrt2cos (x - pi/4)#
Equation (1) becomes:
#sin x + cos x = sqrt2cos (x - pi/4) = 1#
#cos (x - pi/4) = 1/sqrt2#
Trig unit circle and property of cos x function -->
#x - pi/4 = +- pi/4#
a. #x - pi/4 = pi/4#
#x = pi/4 + pi/4 = pi/2#
This answer is rejected because of the above condition: #x != pi/2#
b. #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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