Call t = tan x, we get
(2t)/(1 - t^2)(1/t) - 3 =2t1−t2(1t)−3= (2t)/(t(1 - t^2)) - ((3t)(1 - t^2))/(t(1 - t^2))2tt(1−t2)−(3t)(1−t2)t(1−t2) =
(-t + 3t^3)/(t(1 - t^2)) = (3t^2 - 1)/(1 - t^2) = 0−t+3t3t(1−t2)=3t2−11−t2=0
Conditions: t different to 0, and different to +- 1±1
(3t^2 - 1) = 0 --> t^2 = 1/3 -> t = tan x = +- sqrt3/3(3t2−1)=0−→t2=13→t=tanx=±√33
t = tan x = (sqrt3)/3 -> x = pi/6t=tanx=√33→x=π6
t = tan x = -sqrt3/3 --> x = (5pi)/6t=tanx=−√33−→x=5π6
Check with x = (5pi)/6x=5π6
tan 2x = tan ((10pi)/6) = tan ((5pi)/3) = - tan (pi/3) = - sqrt3tan2x=tan(10π6)=tan(5π3)=−tan(π3)=−√3
cot x = cot ((5pi)/6) = - sqrt3cotx=cot(5π6)=−√3
f(x) = (-sqrt3)(-sqrt3) - 3 = 0f(x)=(−√3)(−√3)−3=0. OK
