Question

If 2tanx/(1-tan^2x)=1/sqrt3 then x can equal:
A) pi/12
B) 7pi/12
C) 11pi/12
D) 5pi/12

Each answer includes nPi multiples, and I understand that tan=pi/3 or tan60, but the values in the answer choice are confusing me.

Answer 1

tan(2x)=tan(pi/6)
x=pi/12

Answer 2

Why is that? I forgot to say it's multiple answer.

Answer 3

\[\frac{2\tan x}{1-\tan ^2x}=\frac{1}{\sqrt{3}}\]

Answer 4

the answer is A

Answer 5

\[2\sqrt{3}\tan x=1-\tan ^2x\]

Answer 6

\[\tan ^2x+2\sqrt{3}\tan x-1=0\]

Answer 7

2tanx/(1-tan^2x)=tan(2x)

That's the double-angle Identity for tangent. So the equation is:

\[\tan(2x)=\frac{ 1 }{\sqrt{3} }\]

So now you need to solve with methods for multiple angle equations... do you know what I mean by that?

Answer 8

@DebbieG That's where I get stuck. I only know how to get the base answers, not the multiple angles.

Answer 9

Also, part of the problem might be that you say above:

"tan=pi/3"

By which I presume you mean to say that \[\tan(\frac{ \pi }{ 3 })=\frac{ 1 }{\sqrt{3} }\]....... you might want to double check that?? :)

Answer 10

OK, here's the key with MA equations: PUT BLINDERS ON. :) In other words, pay no attention to the fact that it's a multiple angle just yet. We'll get to that soon enough. Just supposed you had

\[\tan(x)=\frac{ 1 }{\sqrt{3} }\]
What would be the solution? (in radians, since your answers are in radians)

Answer 11

I would think it was \[\pi/3\] but my foundation of this isn't too swell.

Answer 12

You should know this triangle like the back of your hand. :) And although I did it in degrees, obviously it works for the equivalent radian measures.

Answer 13

So wait... Is it pi/6? I said pi/3 because I divided sin by cos for that point on the unit circle and thought it was 1/sqrt3... I'm going about this the wrong way, huh? :)

Answer 14

Now, that would work, but sin(pi/6)=1/2, cos(pi/6)=sqrt(3)/2

They are opposite for pi/3, so you get tan(pi/3)=sqrt(3)

Personally I prefer the triangle, but the unit circle will work if you use the right values. :)

Answer 15

OK, so if we have an equation like\[\tan(x)=\frac{ 1 }{\sqrt{3} }\]then we know that \[x=\frac{ \pi }{6 }\]is one solution. But here, it isn't x that is in he function, it's 2x... right?

So that means the solution we found is a solution for 2x, not for x. That is:
\[2x=\frac{ \pi }{ 6 }\]right? Understand so far?

Answer 16

Now as you said, we need to add on the multiples of the period, to capture all the coterminal solutions. Now is the time for that:

\[\large 2x=\frac{ \pi }{ 6 }+k\pi\]

Answer 17

Now all that's left is to solve THIS for x.... but don't forget to distribute on the RHS when you do so! Thus, the fact that it's a multiple angle will affect not just the solution, but the increment between solutions - the kpi. Follow me? So you said above that "Each answer includes nPi" and that's normally true for a tangent equation... but since it's a multiple angle, it will NOT be nPi here when you're all finished.

Answer 18

So multiply pi/6 times 2? I'm having difficulty getting the numbers in front pi in the answers. (Ex: The '5' in 5pi/12). How do any relate to pi/6?

Answer 19

Look at this again:

\[\large 2x=\frac{ \pi }{ 6 }+k\pi\]

What do you need to do, in order to solve it for x? Tell me what the operation is....

Answer 20

Oops, wait... Divide by 2?

Answer 21

There ya go... :)

Answer 22

Phew, online class overload, sorry. :) So, okay. Pi/12. Are there others? How are those calculated?

Answer 23

Just the ones that come from the periodicity of the function. That's what I was talking about above - but those are captured by the +kPi, which is now +kPi/2, because of the division due to the multiple angle. Follow?

That's why I said early on, "put blinders on" - you need to forget about the multiple angle, until you have the whole "general form" solution, including the +kPi (I call it "the incrementor" since it accounts for the increment between solutions, but that's my own word, don't like for it in a trig book, lol).

So if this wasn't a multiple choice question, you would get as your complete solution set:

\[\left\{ x|x=\frac{ \pi }{ 12 }+k \cdot\frac{ \pi }{ 2 } \right\}\]or a bit less formally, just:\[x=\frac{ \pi }{ 12 }+k \cdot\frac{ \pi }{ 2 } \]

Answer 24

Okay, I think I finally understand up to that point.

Answer 25

I should add, IF this was a sine or cosine multiple angle equation, you WOULD have an additional piece to the solution set, because there would be a second solution that you have to find. Then add the incrementor to both. The difference here is that tangent has a period of pi, rather than 2pi for sine and cosine, so you only need the 1 solution, the incrementor takes care of the rest of it.

Answer 26

I remember learning sine and cosine, but this course never went over the more complex side of tan. That's why I'm kinda clueless for this equation.