My question states: cos(3x)+cos(x)=0

I got two answer pi/2 and pi/4 but there are supposed to be more, can you show me how to get all the answers?

Thanks alot

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- May 8th 2006, 02:06 PMaussiekid90Trig Product Formula Problem
My question states: cos(3x)+cos(x)=0

I got two answer pi/2 and pi/4 but there are supposed to be more, can you show me how to get all the answers?

Thanks alot - May 8th 2006, 02:12 PMaussiekid90Solving for Interval [0, 2pi)
This problem was pretty tough for me:

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

can you show me how to do this? Thanks!

-Aussiekid90 - May 8th 2006, 02:33 PMtopsquarkQuote:

Originally Posted by**aussiekid90**

Thus

This is a quadratic. Let y = cos(x):

Using the quadratic formula I get:

Thus

Now. so we need to take the positive sign. As it happens the result doesn't seem to have a nice form, so:

. Since this angle is second quadrant, we also get an angle .

-Dan - May 8th 2006, 02:47 PMThePerfectHackerQuote:

Originally Posted by**aussiekid90**

Thus,

Thus,

Thus,

Thus, set factors equal to zero.

- May 8th 2006, 02:58 PMThePerfectHacker
You already asked that question in "Urget Math". DO NOT MAKE DOUBLE POSTS

-=USER WARNED=-

The thread has been merged. - May 8th 2006, 03:07 PMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

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

and the other is:

cos(3x)+cos(x)=0

They're close, but not quite the same.

-Dan