Solve from [0,2pi).

Question

cot^2(4x)tan(2x)-tan(2x)=0

How do you solve this (step-by-step)?

Answer 1

first take out a common factor
tan2x(cot^2(4x) - 1) = 0
tan2x(cot4x + 1)(cot4x - 1) = 0

tan2x = 0 OR cot4x = -1 OR cot4x = 1

so if
tan2x = 0
2x = 0 or pi
x = 0 or pi/2

if cot4x = 1
tan 4x = 1
4x = pi/4 or 4x = 5pi/4
x = pi/16 or x = 5pi/16

if cot4x = -1
4x = pi-pi/4 or 4x = 2pi-pi/4
x = 3pi/16 or x = 7pi/16

so x = 0, pi/16, 3pi/16, 5pi/16, 7pi/16

Answer 2

Thank you so much!