How do you find the x intercepts of #y=tan^2((pix)/6)-3#?

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Answer 1 · 2017-03-18T07:53:21

#x={.........-16,-14,-10,-8,-4,-2,2,4,8,10,14,16,.........}#

#y = tan^2((pix)/6) -3#

#x#-intercept is when #y = 0#

#0 = tan^2((pix)/6) - 3#

or #3 = tan^2((pix)/6)#

#tan((pix)/6)=+-sqrt3#

#tan^-1(+-sqrt3) = (pix)/6# and as #tan(+-pi/3)=+-sqrt3#

#(pix)/6=npi+-pi/3#,, where #n# is an integer

#x/6=n+-1/3#

#x=6n+-2#

i.e. #x={.........-16,-14,-10,-8,-4,-2,2,4,8,10,14,16,.........}#

graph{(tan(pix/6))^2-3 [-16, 16, -8, 8]}