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

2 Answers
Nghi N
Feb 24, 2017

#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#

Manikandan S.
Feb 24, 2017

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 #

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