Solution in the interval

Can someone please help me with this problem? I am completely lost on this and I haven't been doing so well in this class at all. If someone could explain how to do this I would really appreciate it. Thank you

Find all solutions in the interval [0, 2π).

(sin x)(cos x) = 0

This equation will be true for:

a) sin(x) = 0

b) cos(x) = 0

For what values of x in the given interval are either true?

How do I find out? Besides punching in every number between 0 and 2pi?

Thanks for answering btw.

These are among the special angles which you should have memorized.

For sin(x) = 0, these represent points on the unit circle whose y-coordinate is zero, that is, are on the x-axis. What angles correspond to these points?

For cos(x) = 0, these represent points on the unit circle whose x-coordinate is zero, that is, are on the y-axis. What angles correspond to these points?

Quote:

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Originally Posted by Darion

How do I find out? Besides punching in every number between 0 and 2pi?

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You are suppose to know that already.
if

if

Oh, Ok. So then sin(x) would be 0 and pi while cos(x) would be pi/2 and 3pi/2

Yes, that's right. :)

Thank you so much guys.

So if it were cos x = sin x, would the answer be pi/4 and 5pi/4?

Yes, that is correct as well!

Thank you so much. You guys are awesome.

I have just one last question if you're willing to help

Find all solutions to the equation.

cos2x + 2 cos x + 1 = 0



I assume this is about the same concept as the other two questions. But I don't know how to set it up.

Quote:

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Originally Posted by Darion

Find all solutions to the equation.
cos2x + 2 cos x + 1 = 0

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Please learn to post in correct notation.
Is it or

cos^2(x)+2cos(x)+1=0

Quote:

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Originally Posted by Darion

cos^2(x)+2cos(x)+1=0

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Then it becomes or

Hint: use a double-angle identity for cosine, so that the equation becomes a quadratic in cos(x).

edit: Nevermind, I interpreted your equation incorrectly.

I'm sorry, I don't know how to use that.