How do you solve #sin3 theta = -1# if #(0 < theta < 2pi)#?

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Answer 1 · 2016-04-18T16:01:06

#S={pi/2, (7pi)/6, (11pi)/6}#

#3theta=sin^-1 (-1)#

#3theta=(3pi)/2 +2pin#

#theta=(3pi)/2 xx 1/3 +2pin xx 1/3#

#theta=pi/2+(2pi)/3 n#

Now put in n values. Remember n values are integers.
#n=0, x=pi/2#

#n=1, x=(7pi)/6#

#n=2, x=(11pi)/6#

#n=3,x=(5pi)/2 > 360# so we stop

#S={pi/2, (7pi)/6, (11pi)/6}#