Solution in the interval

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• Nov 9th 2012, 10:57 AM
Darion
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
• Nov 9th 2012, 11:02 AM
MarkFL
Re: Solution in the interval
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?
• Nov 9th 2012, 11:15 AM
Darion
Re: Solution in the interval
How do I find out? Besides punching in every number between 0 and 2pi?

• Nov 9th 2012, 11:23 AM
MarkFL
Re: Solution in the interval
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?
• Nov 9th 2012, 11:27 AM
Plato
Re: Solution in the interval
Quote:

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

You are suppose to know that already.
$\displaystyle \sin(t)=0$ if $\displaystyle t=0\text{ or }t=\pi~.$

$\displaystyle \cos(t)=0$ if $\displaystyle t=\frac{\pi}{2}\text{ or }t=\frac{3\pi}{2}~.$
• Nov 9th 2012, 11:41 AM
Darion
Re: Solution in the interval
Oh, Ok. So then sin(x) would be 0 and pi while cos(x) would be pi/2 and 3pi/2
• Nov 9th 2012, 11:42 AM
MarkFL
Re: Solution in the interval
Yes, that's right. :)
• Nov 9th 2012, 11:50 AM
Darion
Re: Solution in the interval
Thank you so much guys.

So if it were cos x = sin x, would the answer be pi/4 and 5pi/4?
• Nov 9th 2012, 11:53 AM
MarkFL
Re: Solution in the interval
Yes, that is correct as well!
• Nov 9th 2012, 12:08 PM
Darion
Re: Solution in the interval
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.
• Nov 9th 2012, 12:15 PM
Plato
Re: Solution in the interval
Quote:

Originally Posted by Darion
Find all solutions to the equation.
cos2x + 2 cos x + 1 = 0

Please learn to post in correct notation.
Is it $\displaystyle \cos(2x)+2\cos(x)+1=0$ or $\displaystyle \cos^2(x)+2\cos(x)+1=0~?$
• Nov 9th 2012, 12:22 PM
Darion
Re: Solution in the interval
cos^2(x)+2cos(x)+1=0
• Nov 9th 2012, 12:25 PM
Plato
Re: Solution in the interval
Quote:

Originally Posted by Darion
cos^2(x)+2cos(x)+1=0

Then it becomes $\displaystyle (\cos(x)+1)^2=0$ or $\displaystyle \cos(x)=-1.$
• Nov 9th 2012, 12:32 PM
MarkFL
Re: Solution in the interval
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.
• Nov 9th 2012, 03:12 PM
Darion
Re: Solution in the interval
I'm sorry, I don't know how to use that.
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