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

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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:

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

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

How do you solve 4\sin ^ { 2} 3x - 1= 04sin23x−1=04\sin ^ { 2} 3x - 1= 0?