# Thread: Solution in the interval

1. ## 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

2. ## 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?

3. ## Re: Solution in the interval

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

4. ## 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?

5. ## Re: Solution in the interval

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.
$\sin(t)=0$ if $t=0\text{ or }t=\pi~.$

$\cos(t)=0$ if $t=\frac{\pi}{2}\text{ or }t=\frac{3\pi}{2}~.$

6. ## 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

7. ## Re: Solution in the interval

Yes, that's right.

8. ## 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?

9. ## Re: Solution in the interval

Yes, that is correct as well!

10. ## 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.

11. ## Re: Solution in the interval

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 $\cos(2x)+2\cos(x)+1=0$ or $\cos^2(x)+2\cos(x)+1=0~?$

12. ## Re: Solution in the interval

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

13. ## Re: Solution in the interval

Originally Posted by Darion
cos^2(x)+2cos(x)+1=0
Then it becomes $(\cos(x)+1)^2=0$ or $\cos(x)=-1.$

14. ## 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.

15. ## Re: Solution in the interval

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

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