Question

How do you solve `tan2x = 8cos^2x - cotx` from `[0,pi/2]`?

Answer 1

Solve `tan 2x = 8cos^2 x - cot x` `[0, pi/2)`

Explanation:

Call tan x = t
Replace `tan 2t = (2t)/(1 - t^2)` and `cos^2x = 1/(1 + t^2)`
`f(t) = (2t)/(1 - t^2) - 8/(1 + t^2) + 1/t = 0`
`2t^2(1 + t^2) - 8t(1 - t^2) + (1 - t^4) = 0`
Solve this equation in the 4th power for t:
`t^4 + 8t^3 + 2t^2 - 8t + 1 = 0`
Use graphing calculator to find t = tan x that gives the answer
`0 < x < pi/2`
I need some help for accuracy.
I found inside interval (0, pi/2) two values of t:
t = tan x = 0.13 --> x = 7.40 deg
and t = tan x = 0.77 --> x = 37.60 deg
Check. x = 7.40 -> tan 2x = tan 14.8 = 0.26. -> 8cos ^2 (7.40) = 7.86 --> cot (7.40) = 1/tan (7.40) = 7.60 -->
0.26 = 7.85 - 7.60 OK
Check. x = 37.69 --> tan 2x = tan 3.78 --> 8cos^2 x = 5.02 --> cot x = 1.30 --> 3.78 = 5.02 - 1.30 OK.