How do you solve the equation on the interval [0,2pi) for #|tan(t)|=1/sqrt(3)#?

Question

10203 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-29T05:57:26

#pi/3, (2pi)/3, (4pi)/3 and (5pi)/3#

#|tan theta|=1/sqrt 3# is the combined equation for the pair

#tan theta = 1/sqrt 3 and tan theta = -1/sqrt 3#

The principal values of theta are #+-pi/3#.

The general values are #npi+pi/3, n= 0, +-1, +-2,...# and

#npi-pi/3, n= 0, +-1, +-2,...#

In #(0, 2pi)#, the first gives #pi/3 (n=0) and (4pi)/3(n=1).

In #(0, 2pi)#, the second gives #2pi/3 (n=1) and (5pi)/3(n=21).