More Trigofun!

[The Hobo]

Having a bit of a math problem, here...

I have this problem that I need to solve:

sin(2x) - cos(x) = 0

So I changed it to this:

sin(x)cos(x) + cos(x)sin(x) - cos(x) = 0

But now I really don't know where to go from there.

Any clues for me?

[bugzpodder]

do you see that sin(x)cos(x)+cos(x)sin(x)=2sin(x)cos(x)?

now you have 2sin(x)cos(x)-cos(x)=0

you can factor out a cos(x)

cos(x)(2sin(x)-1)=0

this is like (x-3)(x-5)=0 can you take it from here?

[The Hobo]

Oh, factoring! I hadn't thought of that (obviously).

I'll mess around and see what I can come up with. And if I can't come up with anything, I'll just look at your post again and...yeah.

Thanks, man.

[The Hobo]

so...

2sin(x)cos(x) - cos(x) = 0
cos(x)(2sin(x) - 1) = 0

cos(x) = 0
x = pi/2 ...


2sin(x) - 1 = 0
sin(x) = 1/2

x = pi/6, 5pi/6

I'm lost with the cos(x) = 0...is 0 positive or negative? Do I list values for all quadrants?

[The Hobo]

This is going up to 2pi, by the way. Should have mentioned that.

[Kalkewl8ter]

cos(x)=0
x=pi/2+2*pi*k or x=(3/2)*pi+2*pi*k
sin(x)=1/2
x=pi/6+2*pi*k or x=(5/6)*pi+2*pi*k,
where k can take on any integral value.

[The Hobo]

I know.

Obviously I'm having a few problems executing that for cos(x) = 0 because I don't know what Quadrants to do it for, as I stated.

[bugzpodder]

but it doesn't matter what quadrant right? you are looking for values of x from 0 to 2pi you said? then cosx=0 gives you x=pi/2 and 3pi/2

[Kalkewl8ter]

Where does it say 0 to 2*pi?

[The Hobo]

Originally posted by Kalkewl8ter
Where does it say 0 to 2*pi?

On my assignment.

Originally posted by bugzpodder
but it doesn't matter what quadrant right? you are looking for values of x from 0 to 2pi you said? then cosx=0 gives you x=pi/2 and 3pi/2

But cos(5pi/2) also = 0. Why wouldn't that be one of them?

[The Hobo]

Oh...because it's bigger than 2pi...duh

[bugzpodder]

Originally posted by The Hobo
This is going up to 2pi, by the way. Should have mentioned that.

right there