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Mathematics 16 Online
OpenStudy (anonymous):

Find all solutions to the equation. (sin x)(cos x) = 0

OpenStudy (cwrw238):

either sin x = 0 or cos x 0

OpenStudy (anonymous):

this is due to the zero property of multiplication

OpenStudy (anonymous):

These are the options they gave me:

OpenStudy (anonymous):

Would it be the first one?

OpenStudy (jdoe0001):

so, at what points are sin(x) = 0 and cos(x) = 0?

OpenStudy (jdoe0001):

check your Unit Circle, if you don't have one, you'd need one, many online

OpenStudy (anonymous):

It would be the second one then right?

OpenStudy (anonymous):

Because 0 is at 0, pi/2, pi, and 3pi/2

OpenStudy (jdoe0001):

yes, the notation pretty much is just saying, every \(\large \frac{\pi}{2}\) increment onwards

OpenStudy (jdoe0001):

and yes, at ever \(\frac{\pi}{2}\) either sine or cosine will zero out

OpenStudy (anonymous):

Or would it be the 4th one because that one says every pi/2

OpenStudy (jdoe0001):

those 2 look identical to me :/

OpenStudy (anonymous):

The second option is pi/2 + n pi & n pi The fourth one is just pi/2 + n pi

OpenStudy (jdoe0001):

yes, but the restrictions application seem to be exactly the same for the exact same elements

OpenStudy (anonymous):

Which do you think would be a better answer then?

OpenStudy (jdoe0001):

the 2nd one is just more redundant

OpenStudy (anonymous):

I think the second one instead of the fourth but I'm not sure

OpenStudy (jdoe0001):

I see nothing wrong with either, but I guess go for the redundant one heheh

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