How do you solve for cos2x = 2 cosx sinx?

  • Thread starter Maria
  • Start date
In summary, the conversation is about finding the values of x for which the equation cos2x = 2 cosx sinx holds. It is suggested to use the fact that 2cos(x)sin(x) = sin(2x) to transform the equation into something easier. The equation is then transformed into sin(2x) = 2cos(x)sin(x) and it is explained that this is an identity and is true for all values of x. The conversation then moves on to finding four angles that satisfy the equation cos2x = 2 cosx sinx. It is explained that the four angles are obtained by setting tan(2x) = 1 and solving for x using the equation tan(2x
  • #1
Maria
Can someone please walk me trough this one:
cos2x = 2 cosx sinx :uhh:
 
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  • #2
You are to find those x for which the equation holds (right?)
What have you thought of thus far?
 
  • #3
you might use the fact that 2cos(x)sin(x) = sin(2x) to transform the equation into something easier.
 
  • #4
mathwonk said:
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.. :wink:
Seems like I scared the poster off instead by my questioning..
 
  • #5
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.
 
  • #6
I need to find 4 angles..
I have found that cos2x = sin2x
correct?
But I am not sure why?
 
  • #7
Oh sorry.. Mathwonk..I understand.. stupid me..
 
  • #8
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 ?
 
  • #9
You do have cos2x=sin2x
Dividing with cos2x, you get:
tan2x=1.
 
  • #10
of course.. forgot..
then i get
x =22,5
this is one angle..
 
  • #11
Certainly; how would you find the others (when restricting 0<x<360)
 
  • #12
180+22,5 = 202,5

but how about the last two?
 
  • #13
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
 
  • #14
So I can for instance set n=1,but thenI get x = 113
am I right?
 
  • #15
No, you get for n=1: x=22.5+90=112.5
 
  • #16
and for n=-1 I get 292,5
 
  • #17
How du you get from tan(y+180n)=tan(y) to the equation:
2x=45+180n?
 
  • #18
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.
 
  • #19
so I don`t need tan since I have it on both sides?
 
  • #20
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)).
 
  • #21
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?
 
  • #22
Ok, I see what your getting at (I can be a bit slow..:redface:)
Yes, I would say you could do that.
 
  • #23
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
 
  • #24
In order to get the 4 values of x lying between 0 and 360, you should use
n=0,1,2,3
 
  • #25
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?
 
  • #26
When I see your setup, I think you might simplify as follows:
...
tan(2x)=1
which implies:
2x=45+180n, for some integer n
...
 
  • #27
I see.
but everything else is ok?
 
  • #28
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. :biggrin:
 
Last edited:
  • #29
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
 
  • #30
well, she doesn`t like short answers, but not to long either
 
  • #31
I have on stupid question left:
Why do I get 4 angles instead og just 2?
Is it because tan =1?
 
  • #32
You get 4 distinct angles because you must solve tan(2x)=1 rather than tan(x)=1
If for example, you were to solve tan(4x)=1, you would have even more distinct solutions
(You could work out how many for yourself)
 
  • #33
thanks a lot for answering all my stupid questions...
 

1. How do you solve for cos2x = 2 cosx sinx?

To solve for cos2x = 2 cosx sinx, you can use the double angle formula for cosine: cos2x = cos^2x - sin^2x. Then, substitute 2 cosx sinx for cos2x to get the equation 2 cosx sinx = cos^2x - sin^2x. From there, you can use algebraic manipulations to solve for x.

2. What is the double angle formula for cosine?

The double angle formula for cosine is cos2x = cos^2x - sin^2x.

3. Can I use a calculator to solve for cos2x = 2 cosx sinx?

Yes, you can use a calculator to solve for cos2x = 2 cosx sinx. However, it is important to note that calculators may not always give exact solutions, so it is important to double check your answer using algebraic manipulations.

4. Are there any other ways to solve for cos2x = 2 cosx sinx?

Yes, there are other ways to solve for cos2x = 2 cosx sinx. One method is to use the half angle formula for cosine: cos2x = 2 cos^2(x/2) - 1. Another method is to use the Pythagorean identity: cos^2x + sin^2x = 1. By rearranging this identity, you can also solve for cos2x = 2 cosx sinx.

5. Can this equation be solved for any value of x?

Yes, this equation can be solved for any value of x. However, it is important to note that there may be multiple solutions or no solutions at all, depending on the values of cosx and sinx. It is important to carefully consider the domain and range of the trigonometric functions when solving for x.

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