How do you solve #4\sin ^ { 2} 3x - 1= 0#?

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Answer 1 · 2017-11-24T14:45:33

#x= (+-pi)/18+(2kpi)/3 or (+-5pi)/18+(2kpi)/3, k in NN#

#4sin^2 3x−1=0#

This is an equation of the type

#4y^2-1=0#

So we must start from that:

#4y^2-1=0#

#(2y+1)(2y-1)=0#

So, # y=+-1/2#

Remember #->y=sin 3x#

#sin3x=+-1/2#

See in the trigonometrical circle and realize that:

#3x= +-pi/6+2kpi or (+-5pi)/6+2kpi, k in NN#

#x= (+-pi)/18+(2kpi)/3 or (+-5pi)/18+(2kpi)/3, k in NN#