Solve this: $| 2 cos 3 x | = 1$ $abs(2cos3x)=1$ ---> (x=... + ...)?

Question

$| 2 cos 3 x | = 1$ $abs(2cos3x)=1$

Will it be: $x = \frac{π}{9} + \frac{2}{3} k π$ $x=pi/9+2/3kpi$ AND $x = - \frac{π}{9} + \frac{2}{3} k π$ $x=-pi/9+2/3kpi$
OR ......
$x = \frac{π}{9} + \frac{1}{3} k π$ $x=pi/9+1/3kpi$ AND $x = - \frac{π}{9} + \frac{1}{3} k π$ $x=-pi/9+1/3kpi$

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Answer 1 · 2017-04-02T15:54:45

$x = \frac{2}{3} k π \pm \frac{π}{9}$ $x=2/3kpi+-pi/9$ and $x = \frac{2}{3} k π \pm \frac{2 π}{9}$ $x=2/3kpi+-(2pi)/9$

As $| 2 cos 3 x | = 1$ $|2cos3x|=1$ , we have

either $2 cos 3 x = 1$ $2cos3x=1$ i.e. $cos 3 x = \frac{1}{2} = cos (\frac{π}{3})$ $cos3x=1/2=cos(pi/3)$

and $3 x = 2 k π \pm \frac{π}{3}$ $3x=2kpi+-pi/3$ or $x = \frac{2}{3} k π \pm \frac{π}{9}$ $x=2/3kpi+-pi/9$

or $2 cos 3 x = - 1$ $2cos3x=-1$ i.e. $cos 3 x = - \frac{1}{2} = cos (\frac{2 π}{3})$ $cos3x=-1/2=cos((2pi)/3)$

and $3 x = 2 k π \pm \frac{2 π}{3}$ $3x=2kpi+-(2pi)/3$ or $x = \frac{2}{3} k π \pm \frac{2 π}{9}$ $x=2/3kpi+-(2pi)/9$