Use trigonometric identities to solve the equation csc^2x -(√3+1)cotx+(√3-1)=0 in the interval [0, 2pi]

Answer 1

One of the ID's you should know is
cot^ Ø + 1= csc^ Ø

then:

csc^2x -(√3+1)cotx+(√3-1)=0
cot^2 x + 1 - (√3+1)cotx + (√3-1)=0
cot^2 x - (√3+1) cotx + √3 = 0
cot^2 x - cotx - √3cotx + √3=0
cotx(cotx - 1) - √3(cotx -1) = 0
(cotx - 1)(cotx - √#) = 0

cotx = 1 or cotx = √3

case1: cotx = 1
then tanx = 1
x = 45°, 225° or π/4 , 5π/4 radians

case 2: cotx = √3
tanx = 1/√3
recognize the 1-√3-2 right-angled triangle,
x = 30° , 210° or π/6, 7π/6

x = π/6, π/4, 7π/6 , 5π/4