How do I solve #sqrt3 tan3theta + 1 = 0#?

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Answer 1 · 2018-05-07T01:06:16

Within the interval #0 < theta < 2pi:#

#theta=(5pi)/18, (11pi)/18#

.

#sqrt3tan3theta+1=0#

#tan3theta=-1/sqrt3=-sqrt3/3#

#3theta=arctan(-sqrt3/3)#

#3theta=(5pi)/6,(11pi)/6#

Within the interval #0 < theta < 2pi:#

#theta=(5pi)/18, (11pi)/18#

Answer 2 · 2018-05-07T01:08:48

#theta= (5pi)/18+pi/3n#

#sqrt3 tan3theta + 1 = 0#

Let #u# be #3theta#

#sqrt3 tanu + 1 = 0#

#sqrt3 tanu = -1#

#tanu = -1/sqrt3#

#tanu = -sqrt3/3#

#u= (5pi)/6, (11pi)/6#

So:
#3theta=(5pi)/6 +pin#

#theta= (5pi)/18+pi/3n#
#n# is an element of all integers

graph{sqrt3tan(3x)+1 [-10, 10, -5, 5]}