How do you solve #sec^2x+tanx-1=0#?

1 Answer
Nghi N.
Sep 26, 2016

#0, (3pi)/4, pi, (7pi)/4, 2pi#

Explanation:

There are 2 variables: sin x and cos x. General Method: we must transform the trig equation into a product of 2 simple trig equations.
#1/(cos^2 x) + sin x/(cos x) = 1#
#1 + sin x.cos x = cos^2 x#
#(1 - cos^2 x) + sin x.cos x# = 0
#sin^2 x + sin x.cos x = 0#
#sin x(sin x + cos x) = 0#
Now, we solve the two simple trig equations.
a. sin x = 0 --> #x = 0, and x = pi, and x = 2pi#
b. Use trig identity:
#sin a + cos a = sqrt2cos (a - pi/4)#
#sin x + cos x = sqrt2cos (x - pi/4) = 0#
#cos (x - pi/4) = 0#
Trig unit circle -->
c. #x - pi/4 = pi/2# -->
#x = pi/2 + pi/4 = (3pi)/4#
d. #x - pi/4 = (3pi)/2# -->
#x = (3pi)/2 + pi/4 = (7pi)/4#
Answers for #(0, 2pi)#
#0, (3pi)/4, pi, (7pi)/4, 2pi#

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