How do you graph to solve the equation on the interval #[-2pi,2pi]# for #cotx=-sqrt3/3#?

1 Answer
Aug 5, 2017

Answer:

Solution : In interval #[-2pi,2pi] , x= -pi/3 = -60^0, x = -(4pi)/3=-240^0 x= (2pi)/3=120^0 , x = (5pi)/3=300^0#

Explanation:

#tan (pi/3) =3/sqrt3 :. cot (pi/3) =sqrt (3)/3 or cot (pi-pi/3) = -sqrt (3)/3 #

#cot x = -sqrt (3)/3 :. x = (pi-pi/3) = (2pi)/3 ; x =pi+ (2pi)/3 = (5pi)/3#

#(5pi)/3 =((2pi)-(5pi)/3)= -pi/3 :. x = -pi/3;#

# (2pi)/3 =((2pi)-(2pi)/3)= -(4pi)/3 :. x = -(4pi)/3 #

Solution : In interval #[-2pi,2pi] , x= -pi/3 = -60^0, x = -(4pi)/3=-240^0 x= (2pi)/3=120^0 , x = (5pi)/3=300^0#[Ans]