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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