Question

How do you solve `2cotx-4/cotx=7` in the interval `0<=x<=2pi`?

Answer 1

`14^@04; -63^@43`

Explanation:

`2cot x - 4/(cot x) - 7 = 0`
`2cot^2 x - 7cot x - 4 = 0`
Solve this quadratic equation for cot x.
`D = d^2 = b^2 - 4ac = 49 + 32 = 81` --> `d = +- 9`
There are 2 real roots:
`cot x = -b/(2a) +- d/(2a) = 7/4 +- 9/4`
a. `cot x = 16/4 = 4` --> `tan x = 1/4`
Calculator gives -->
`x = 14^@04`
b. `cot x = -2/4 = -1/2` --> `tan x = - 2`
`x = -63^@43`