Cot^4 x-csc^2=5. [0,2π) Find X?

#cot^4 x-csc^2=5. [0,2π)" Find "x#?

1 Answer
Mar 7, 2018

#x = pi/6, x=(5pi)/5, x = (7pi)/6, and x = (11pi)/6#

Explanation:

Assuming that you mean:

#cot^4(x)-csc^2(x) = 5, 0<=x<2pi#

Substitute #csc^2(x) = 1+cot^2(x)#

#cot^4(x)-(1+cot^2(x)) = 5, 0<=x<2pi#

Distribute the -1:

#cot^4(x)-cot^2(x)-1 = 5, 0<=x<2pi#

Combine like terms:

#cot^4(x)-cot^2(x)-6 = 0, 0<=x<2pi#

Factor:

#(cot^2(x)-3)(cot^2(x)+2)=0, 0<=x<2pi#

We must discard the second factor because it is imaginary:

#cot^2(x) -3=0, 0<=x<2pi#

#cot(x)= sqrt3 and cot(x)=-sqrt3, 0<=x<2pi#

In case you do not have an inverse cotangent function on your calculator, this is the same thing:

#tan(x) = sqrt3/3 and tan(x) = -sqrt3/3, 0<=x<2pi#

#x = tan^-1(sqrt3/3) and x = tan^-1(-sqrt3/3), 0<=x<2pi#

#x = pi/6, x=(5pi)/5, x = (7pi)/6, and x = (11pi)/6#