Question

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)`?

Answer 1

`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`