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

1 Answer
Jan 9, 2018

Answer:

#x={-(5pi)/3, -(2pi)/3, pi/3, (4pi)/3}#.

Explanation:

Since tangent is positive, we know that the angle comes from either QI or QIII.

Since #tan(x) = sqrt(3)# we know that #x# is a #pi/3# angle.

Within #0<x<2pi#, we know that the angles #pi/3# and #4pi/3# have tangents of #sqrt(3)#.

Since we're solving on #-2pi<x<2pi#, we can subtract #2pi# from each value and stay in the necessary interval:

#pi/3-2pi=-(5pi)/3#
#4pi/3-2pi=-(2pi)/3#

So our answers are #x={-(5pi)/3, -(2pi)/3, pi/3, (4pi)/3}#.