How do you use the inverse functions where needed to find all solutions of the equation #sec^2x+tanx-3=0# in the interval #[0,2pi)#?

1 Answer
Nghi N.
Jun 3, 2018

#45^@; 116^@57;225^@; 296^@57#

Explanation:

#1/cos^2 x + tan x - 3 = 0# (1)
Reminder of trig identity:
#cos^2 x = 1/(1 + tan^2 x)# --> #1/cos^2 x = 1 + tan^2 x#
We get from equation (1):
#1 + tan^2 x + tan x - 3 = 0#
#tan^2 x + tan x - 2 = 0#.
Solve this quadratic equation for tan x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
tan x = 1 and #tan x = c/a = - 2#
a. tan x = 1
Trig table and Unit circle give -->
#x = pi/4# and #x = pi + pi/4 = (5pi)/4#
b. tan x = - 2
Calculator and unit circle give -->
#x = - 63^@43#, and #x = -63.43 + 180 = 116^@57#
Note. x = -63.43 is co-terminal to x = 360 - 63.43 = 296.57.
Answers for (0, 360):
#45^@; 116^@57; 225^@; 296^@57#

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