How do you solve $4sin(3x)=2$ in the interval $[0,(4pi)/3]$ ?

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Answer 1 · 2017-02-02T22:35:11

The solutions included in the interval $[0;4/3pi]$ are

$pi/18; 5/18pi; 13/18pi; 17/18pi$

It is:

$sin(3x)=2/4=1/2$

then

$3x=pi/6+2kpi$ or $3x=pi-pi/6+2kpi$

$x=pi/18+2/3kpi$ or $x=5/18pi+2/3kpi$

The solutions included in the interval $[0;4/3pi]$ are therefore:

$pi/18; 5/18pi; 13/18pi; 17/18pi$

How do you solve 4sin(3x)=24sin(3x)=2 in the interval [0,(4pi)/3][0,(4pi)/3]?