How do you solve #tan^2x=1/3# in the interval #0<=x<=2pi#?

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Answer 1 · 2018-04-26T08:10:03

#x=pi/6,(5pi)/6,(7pi)/6and(11pi)/6,where,x in [0,2pi)#

Here,

#tan^2x=1/3=(1/sqrt3)^2, where, 0 <= x <= 2pi#

#=>tanx=+-1/sqrt3#

Now,

#(i) tanx=1/sqrt3 > 0=>I^(st)Quadrant or III^(rd)Quadrant#

#:.x=pi/6to I^(st)Quadrant or#

#x=pi+pi/6=(7pi)/6toIII^(rd) Quadrant#

#(ii)tanx=-1/sqrt3 < 0=>II^(nd)Quadrant or IV^(th)Quadrant#

#:.x=pi-pi/6=(5pi)/6toII^(nd)Quadrant or#

#x=2pi-pi/6=(11pi)/6to IV^(th)Quadrant#

Hence,

#x=pi/6,(5pi)/6,(7pi)/6and(11pi)/6#