Subtract 1, divide by #sqrt3#, and you get:
#tanx = -1/sqrt3#
This is true in multiple spots.
With #sintheta#, you get the same intersection of the x-axis at #pm kpi# where #k# is an integer.
#costheta# is really a #pi/2# phase shift of #sintheta#, so you get the same thing, except it is #pi/2 pm kpi# instead.
The period of #tantheta# is every #pi# since it takes on the domain of both #sin# and #cos# combined. The phase shift doesn't cause anything more than the vertical asymptotes in the graph, seeing as how whenever #sintheta = pm1#, #costheta = 0#.
You can do this:
#-1/sqrt3 = (-1/2)/(sqrt3/2) = -1/cancel(2)*cancel(2)/sqrt3#
Therefore, it happens wherever:
#sintheta = -1/2# AND #costheta = sqrt3/2#
That means:
#theta = -pi/6 pm (kpi)# where #k# is an integer.
#tanx#:
graph{tanx [-6.24, 6.244, -3.12, 3.12]}
