7sin(x)=0, 0<x<2pi

wlc17math:

Hi, I was wondering how to solve 7sin(x)=0, 0<x<2pi for the exact radian values over the given intervals

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2 things will only multiply to give your 0 if one or both of them is 0 to start with.

therefore your problem is sinx=0
This will be true when y = 0 on the unit circle.

That is 0 and pi radians.

7 sin (x) = 0
7 cannot be equal to 0
so that sin (x) = 0
that is for x = 0, pi, 2pi radians.