# Thread: Where did i go wrong?

1. ## Where did i go wrong?

I have a problem that look like this: cos(3x)=1/ sqrt2

I already got the correct anser but my teacher says i have to give a more detailed description of how i solved it and he says that part of calculatón is wrong, this is my calculation:

The unit circle value for the equation above is 40 degrees or pi/4 radian. after finding this i make 3x=a to make it easier. So i figured that because sin(a)=cos(a) i can from this prove that a=pi/4+2*n*pi ...ooor a=-pi/4+2*n*pi

Then we switch back to 3x and in order to get x alone we need to divide all sides by 3 in both solutions, and that leaves us with the final anwser:

x=pi/12+2*n*pi/3

x=-pi/12+2*n*pi/3

What would you recommend i add or do you see anything incorrect??

2. ## Re: Where did i go wrong?

Originally Posted by MathChick
I have a problem that look like this: cos(3x)=1/ sqrt2

I already got the correct anser but my teacher says i have to give a more detailed description of how i solved it and he says that part of calculatón is wrong, this is my calculation:

The unit circle value for the equation above is 40 degrees or pi/4 radian. after finding this i make 3x=a to make it easier. So i figured that because sin(a)=cos(a) i can from this prove that a=pi/4+2*n*pi ...ooor a=-pi/4+2*n*pi

Then we switch back to 3x and in order to get x alone we need to divide all sides by 3 in both solutions, and that leaves us with the final anwser:

x=pi/12+2*n*pi/3

x=-pi/12+2*n*pi/3

What would you recommend i add or do you see anything incorrect??
The only thing I see wrong with above is that $\pi/4$ radians is 45, not 40, degrees.

The answers you give at the bottom are correct.

3. ## Re: Where did i go wrong?

Oh right that was a mistake, but is there anything else that is not enough explained in your opinion?or does it look good ))

4. ## Re: Where did i go wrong?

Originally Posted by MathChick
Oh right that was a mistake, but is there anything else that is not enough explained in your opinion?or does it look good ))
well it would be clearer if you just said that

$3x=\arccos(1/\sqrt{2})=\pm \pi/4 + 2\pi n, ~~n \in \mathbb{Z}$

$x=\pm \pi/12 + 2\pi n/3, ~~n \in \mathbb{Z}$

rather than the the "method" you used to figure out $\cos(\pi/4)=1/\sqrt{2}$

but other than that it looks fine.

5. ## Re: Where did i go wrong?

Originally Posted by MathChick
The unit circle value for the equation above is 40 degrees or pi/4 radian.
I am not sure what "the unit circle value" means, but this sentence is most likely wrong, even if you replace 40 with 45.

Originally Posted by MathChick
So i figured that because sin(a)=cos(a)
Why?

I agree that last part, including the final answer, is correct.

6. ## Re: Where did i go wrong?

Originally Posted by MathChick
I have a problem that look like this: cos(3x)=1/ sqrt2

I already got the correct anser but my teacher says i have to give a more detailed description of how i solved it and he says that part of calculatón is wrong, this is my calculation:

The unit circle value for the equation above is 40 degrees or pi/4 radian. after finding this i make 3x=a to make it easier. So i figured that because sin(a)=cos(a) i can from this prove that a=pi/4+2*n*pi ...ooor a=-pi/4+2*n*pi

Then we switch back to 3x and in order to get x alone we need to divide all sides by 3 in both solutions, and that leaves us with the final anwser:

x=pi/12+2*n*pi/3

x=-pi/12+2*n*pi/3

What would you recommend i add or do you see anything incorrect??
I suspect that what bothered your teacher is this. Your conversion from cos to sin is, as has been pointed out above, unnecessary, but it also is invalid.

Yes $-\ \dfrac{\pi}{2} \le 3x \le \dfrac{\pi}{2}\ and\ cos(3x) = \dfrac{1}{\sqrt{2}} \implies 3x = \pm\ \dfrac{\pi }{4}.$

But $-\ \dfrac{\pi}{2} \le 3x \le \dfrac{\pi}{2}\ and\ sin(3x) = \dfrac{1}{\sqrt{2}} \implies 3x = \dfrac{\pi }{4}\ and\ 3x \ne -\ \dfrac{\pi }{4}.$