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?

Re: Solution in the interval

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

Thanks for answering btw.

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?

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 $\displaystyle t=\frac{\pi}{2}\text{ or }t=\frac{3\pi}{2}~.$

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

Re: Solution in the interval

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?

Re: Solution in the interval

Yes, that is correct as well!

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.

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

Re: Solution in the interval

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 $\displaystyle \cos(x)=-1.$

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.

Re: Solution in the interval

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