Question

How do you solve `2cot^4x-cot^2x-15=0`?

Answer 1

`pi/6 + kpi`
`(5pi)/6 + kpi`

Explanation:

Call `cot^2 x = T`, we get the quadratic equation:
`2T^2 - T - 15 = 0`
`D = d^2 = b^2 - 4ac = 1 + 120 = 121`--> `d = +- 11`
There are 2 real roots:
`T= -b/(2a) +- d/(2a) = 1/4 +- 11/4`
`T = 12/4 = 3`, and
`T = - 10/4 = -5/2` (rejected because T = cot^2 x must be positive)
By definition:
`tan^2 x = 1/(cot^2 x) = 1/T = 1/3` --> `tan x = +- 1/sqrt3 = +- sqrt3/3`
Trig table and unit circle give -->
a. `tan x = sqrt3/3` --> `x = pi/6 + kpi`
b. `tan x = - sqrt3/3` --> `x = (5pi)/6 + kpi`

Answer 2

Substitution

Explanation:

Let `y = cot^2 x`. Then

`2y^2-y-15 = 0`
`y = (1\pmsqrt(1+4*2*15))/4`
`y = 3,-5/2`
`cot^2(x)` cannot be negative as it is a squared value. Hence
`cot^2 x = 3`
`cotx = +-sqrt(3)`
`tanx = +-1/sqrt(3)`
`x = pi/6+npi,(7pi)/6+npi`