# Math Help - Solving for x with tan and cot

1. ## Solving for x with tan and cot

If tan 4(x) = 2cot4(x) , and 0<x<180 (< equals less than or equal to), then x=?

The answer is 13.7, 76.3, 103.7, 166.3 (degrees).

I graphed the two equations. I graphed tan4x and 2/(tan4x) In the graphs, all of these degree measures do show points of intersection. However, I also see other points of intersection between 0 to 180 that are not in the answer set. Is there an explanation for this?

Additionally, my book said to solve for x by first making it 2/(tan4x) = tan4x Once you solve for x this way, the answer is 13.7. Then they say that since it is between 0-180, we get our 4 angle measures from the answer.

I am confused about this part. I understand that 13.7 and 166.3 will be solutions in the first and second quadrants, but why will 76.3 and 103.7 be solutions?

I hope that made sense. Thanks.

2. ## Re: Solving for x with tan and cot

Originally Posted by benny92000
If tan 4(x) = 2cot4(x) , and 0<x<180 (< equals less than or equal to), then x=?

The answer is 13.7, 76.3, 103.7, 166.3 (degrees).

I graphed the two equations. I graphed tan4x and 2/(tan4x) In the graphs, all of these degree measures do show points of intersection. However, I also see other points of intersection between 0 to 180 that are not in the answer set. Is there an explanation for this?

Additionally, my book said to solve for x by first making it 2/(tan4x) = tan4x Once you solve for x this way, the answer is 13.7. Then they say that since it is between 0-180, we get our 4 angle measures from the answer.

I am confused about this part. I understand that 13.7 and 166.3 will be solutions in the first and second quadrants, but why will 76.3 and 103.7 be solutions?

I hope that made sense. Thanks.
you really need to be more careful in posting equations. tan 4(x) is not the way to type tan(4x).

I'm just guessing here, but if the original equation is ...

$\tan(4x) = 2\cot(4x)$

... then I get eight solutions for x in the interval $0 < x < 180^\circ$ by graphing $y = \tan(4x) - 2\cot(4x)$ and looking for zeros.

x = 13.683 , 31.316 , 58.684 , 76.316 , 103.684 , 121.316 , 148.684 , 166.316

3. ## Re: Solving for x with tan and cot

Originally Posted by benny92000
/snip
I am confused about this part. I understand that 13.7 and 166.3 will be solutions in the first and second quadrants, but why will 76.3 and 103.7 be solutions?

I hope that made sense. Thanks./snip
Because we have $tan(4x)$, the graph becomes "compressed", as Skeeter implied. The function repeats four times as many times as it would with $tan(x)$

When solving, we get:

$tan(4x) = 2cot(4x)$

$tan^2(4x)=2$

$tan(4x)=\pm\sqrt{2}$

$4x=arctan(\pm\sqrt{2})$

Now, as you are looking for solutions that satisfy $0, at this stage you need to find the solutions that satisfy $0<4x<720$, which means you need to continue for four loops around the unit circle when looking for solutions. For this reason, I wouldn't recommend dividing by $4$ at this stage. This is where the other solutions come from.

4. ## Re: Solving for x with tan and cot

So there are indeed more than four solutions? My book lists only four in the answer choice. Quacky, I'm still a little confused about how to find all of the solutions. I understand the 13.7 degree solution, and I somewhat understand the 0<4x<720. I would normally graph this and observe the points of intersection). I'm familiar with All of the solutions are 13.7 degrees away from an axis (x or y).

5. ## Re: Solving for x with tan and cot

Well, the obvious solutions are:
$4x=54.732, 125.268$, but these are for $0<\underbrace{4x}<180$ which, after dividing by 4, will give the solutions for $0 and this is not the required range of solutions.

Continuing by adding $180^o$ to each of these gives:

$4x=234.732$, $305.268$ which are the solutions for $180^o<4x<360^o$, and when divided, will give the solutions in the range of $45

...and if you do this again, you'll get:

$4x=414.732, 485.268$, which are solutions for $360<4x<540$, and will, once divided, equate to the solutions for $90

...and once more:

$4x=594.732, 665.268$ which are solutions for $540<4x<720$, or $135 after dividing by $4$.

So:

$4x=54.732, 125.268, 234.732, 305.268, 414.732, 485.268, 594.732, 665.268$

Which are the solutions for $0<4x<720$ and therefore, once you divide all of these by $4$, you will get the $8$ solutions skeeter found for $0, which is the required range of solutions.

6. ## Re: Solving for x with tan and cot

Thanks for working through this with me. Are we adding 180 because that would put the angle in another quadrant where it would have the same sign? And why do we add 180 instead of 360? I remember being told to add 360 at one point. Is my book wrong for only having four of the solutions? Also, since tan^(-1)(-2^(1/2)) (that's inverse tan of negative square root of two) equals -54.735 degrees, why do we start with the value in the second quadrant? Is that because the range we are dealing with is 0-180?

Sorry for so many questions. I've had to teach myself a ton of math concepts for a state math test I'm taking, and I didn't learn hardly any of this when I took Pre-cal.

Edit: When I graphed it for points of intersection, I came to the 8 solutiosn that you guys did, but the book's answer was what was throwing me off. I figured that I had something wrong with the window's settings or equation to get so many solutions.

7. ## Re: Solving for x with tan and cot

The way to check whether the book is wrong is by verifying each of the solutions in the original equation by substituting them into the original equation: check the 4 solutions the book doesn't mention; do they work?

As for why we add $180$, this is based on the periodic motion of the tan function and the unit circle.

$\tan(\alpha)=\tan(\alpha+180)$ and there are a number of algebraic and geometric proofs for this. For a geometric one, (not a proof, but an explanation), you can see clearly by graphing $\tan(x)$ that it is periodic with period $180^o$. Alternatively, you can just look at the unit circle and recall that $\tan(x)$ is positive in the appropriate quadrants, as you suggested, and therefore will have solutions in such quadrants.

We could use the identity $\tan(\alpha +\beta)=\frac{\tan\alpha + \tan\beta}{1- \tan\alpha\cdot\tan\beta}$ with $\beta = 180$ to give:

$\tan(\alpha + 180)=\frac{\tan\alpha + \tan{180}}{1- \tan\alpha \tan{180}}$

and $\tan{180}=0$, so we have:

$\tan(\alpha + 180)=\frac{\tan\alpha + 0}{1- 0}$

$\tan(\alpha + 180)=\tan(\alpha)$

As for the other question, I don't think the quadrant is actually relevant, although I'm not sure I understand what you're asking.