How do you solve for cos2x = 2 cosx sinx?

[Maria]

Can someone please walk me trough this one:
cos2x = 2 cosx sinx

 

[arildno]

You are to find those x for which the equation holds (right?)
What have you thought of thus far?

 

[mathwonk]

you might use the fact that 2cos(x)sin(x) = sin(2x) to transform the equation into something easier.

 

[arildno]

you might use the fact that 2cos(x)sin(x) = sin(2x) to transform the equation into something easier.

I would have gotten to that eventually..
Seems like I scared the poster off instead by my questioning..

 

[HallsofIvy]

Maria, look closely at what Mathwonk said: your equation has cos(2x)= 2cos(x)sin(x)and you are to find the values of x for which it is true. His equation has sin(2x)= 2cos(x)sin(x) and is an IDENTITY: it is true for all values of x.

 

[Maria]

I need to find 4 angles..
I have found that cos2x = sin2x
correct?
But I am not sure why?

 

[Maria]

Oh sorry.. Mathwonk..I understand.. stupid me..

 

[Maria]

I think i`ve got it now
cos2x = 2cosx sonx
cos2x = sin2x
cosx
sinx = 1
tanx = 1

and i get 4 angles because og tan = 1 ?

 

[arildno]

You do have cos2x=sin2x
Dividing with cos2x, you get:
tan2x=1.

 

[Maria]

of course.. forgot..
then i get
x =22,5
this is one angle..

 

[arildno]

Certainly; how would you find the others (when restricting 0<x<360)

 

[Maria]

180+22,5 = 202,5

but how about the last two?

 

[arildno]

Now, remember that tan(y+180)=tan(y)
Hence, for any integer n, we have:
tan(y+180n)=tan(y)
Set y=45 (i.e, so that tan(y)=1), we may find solutions 0<x360
by looking at various choices n in the equation:
2x=45+180n

 

[Maria]

So I can for instance set n=1,but thenI get x = 113
am I right?

 

[arildno]

No, you get for n=1: x=22.5+90=112.5

 

[Maria]

and for n=-1 I get 292,5

 

[Maria]

How du you get from tan(y+180n)=tan(y) to the equation:
2x=45+180n?

 

[arildno]

Well, you're seeking x-solutions satisfying
tan(2x)=1, or hence:
tan(2x)=tan(45+180n) for some n
By setting 2x=45+180n, you're guaranteed the last equation is fulfilled.

 

[Maria]

so I don`t need tan since I have it on both sides?

 

[arildno]

I hope you have accepted that the x-solutions you're looking for must satisfy:
tan(2x)=tan(45+180n), where n is some integer (We call this equation (e)).
Now, to guarantee that (e) holds , requiring 2x=45+180n is evidently enough, since the lefthand side term of (e) (that is tan(2x)) becomes necessarily equal to the righthandside term in (e) (that is tan(45+180n)).

 

[Maria]

I understand that both sides have to be equal. But is it ok if I write

tan(2x) = tan(45+180n)
2x = 45+180n
I don`t need to write anything else in between?

 

[arildno]

Ok, I see what your getting at (I can be a bit slow..)
Yes, I would say you could do that.

 

[Maria]

If I can do that, then I think I get it..
If I set n=1 I get x=112,5 and
if I set n=-1 I get 2x= -135
x = -67,5 which give an angle 360-67,5 = 292,5

 

[arildno]

In order to get the 4 values of x lying between 0 and 360, you should use
n=0,1,2,3

 

[Maria]

you`re right...
I can write it like this:
cos2x = 2cosx sinx
cos2x = sin2x

tan 2x = 1
which gives

tan(2x) = tan(45+180n)
2x = 45+180n

I use n=0,1,2,3 and get the angles
x = 22,5 ^ 112,5 ^ 202,5 ^ 292,5

forgot something?

 

[arildno]

When I see your setup, I think you might simplify as follows:
...
tan(2x)=1
which implies:
2x=45+180n, for some integer n
...

 

[Maria]

I see.
but everything else is ok?

 

[kronecker]

go straig like this is ok 2
cos2x=sin2x;
<=>sin2x-cos2x=0;
<=>sin(2x-Pi/4)=0;
=>2x-Pi/4=k2Pi (k=integer);
chose 4 values of k to get x
your loking for x, so dividing both equation by cos like what you have did isn't good practise.

 

[arildno]

Sure.
It depends a bit on how your teacher likes your answers.
For example, you might insert short explanations between your steps, for example:
...
cos(2x)=sin(2x)
Dividing this equation with cos(2x), we get:
tan(2x)=1

 

[Maria]

well, she doesn`t like short answers, but not to long either