Weird trig discrepancy? (1 Viewer)

[kev-]

tan(2x)=cot(x) - Find the solutions of x for 0 <= x <= 2pi

What I did was double angle for tan, and convert cot to 1/tanx

Then solving for tan x from the quadratic, I got tanx = +- pi/6 + kpi

However, when subbing x=pi/2 into the question, LHS=RHS but it is not one of the solutions from the quadratic.

Why is this?

 

[Elise8842]

What is the problem? Don't we just solve it normally?


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[Elise8842]

Ermmm...a slip of pen in the last line, but you get the idea~~~


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[kev-]

The problem is when I sub in x=pi/2 into tan2x=cotx, it yields true. However, when solving the tanx= +- 1/rt3, x=pi/2 is not one of the solutions. I am wondering why not.

Thanks for replying

 

[kev-]

bump

 

[integral95]

(where k is an integer) is undefined, and the quadratic equation you got is in terms of tan x so it's not possible to obtain solutions even though it's a solution to the original equation.

 

[kev-]

Does that mean my method is flawed somehow?

If I change the cot x into tan(pi/2 - x), x=pi/2 comes out.

What is the difference between these two methods and why does one have an extra solution?

 

[integral95]

Yeah because like i said you got an equation in terms of tan x. so x can't be pi/2 in those lines as it's undefined.

I'm assuming you used general solution to get which would generate x = pi/2

, or how else did you get x = pi/2?

 

[kev-]

Yep, I used general solutions.

I get what you're saying, but does this mean my method is wrong as it doesn't have all the possible solutions?

thanks

 

[anomalousdecay]

http://img.tapatalk.com/d/14/06/20/y7uba7ab.jpg[IMG]
What is the problem? Don't we just solve it normally?


~[/QUOTE]

Those square roots look so cool (but a bit confusing).

[quote="MethewYan, post: 6646965"]Yeah because like i said you got an equation in terms of tan x. so x can't be pi/2 in those lines as it's undefined.

I'm assuming you used general solution to get [tex] x = \frac{2k\pi+\pi}{6} [/tex] which would generate x = pi/2

, or how else did you get x = pi/2?[/QUOTE]

AD to the rescue :p


If you look carefully you should get something like this:

[tex] \frac{2tan(x)}{1- tan^2(x)} - \frac{1}{tan(x)} = 0 [/tex]


So when you multiply by tan(x), how do we know you aren't multiplying by zero or infinity here?

If tan(x) --> infinity here, then you have to make sure that that you aren't solving the equation with :

[tex] 0$ x $ \infty [/tex]

because if you are then you are doing it all wrong.

So you have to account for this situation.

Best way to do that is expand and simplify to get:

[tex] \frac{3tan^2(x) - 1}{tan(x) - tan^3(x)} = 0[/tex]

Now from here clearly as x --> pi/2, we have solutions.

Therefore the solutions are:

[tex] x = \pm \frac{\pi}{6} + k\pi $ and $ x = \frac{\pi}{2} + k\pi [/tex]

where k is any integer.

You have to be careful with your manipulations because you can lose solutions if you just cross multiply it.

Generally you have to be careful with fractions involving solving trig equations.

 

[kev-]

I get it, thank you!

 

[funnytomato]

tan(2x)=cot(x) - Find the solutions of x for 0 <= x <= 2pi

What I did was double angle for tan, and convert cot to 1/tanx

Then solving for tan x from the quadratic, I got tanx = +- pi/6 + kpi

However, when subbing x=pi/2 into the question, LHS=RHS but it is not one of the solutions from the quadratic.

Why is this?

whenever you use t-results , you ALWAYS have to check that if x= pi/2 is a solution precisely because tan(pi/2) is undefined (as MethewYan mentioned before)

alternatively the argument using 'general solutions' might be easier

 

[BLIT2014]

What is the problem? Don't we just solve it normally?


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You have really interesting hand writing