# Determine the general solution for the equation tan 2x=1all I know is this!period=pi/b   or  b=pi/periodgeneral solution: X=angle of first positive asymptote +-n(period)The general equation of asymptotes is: x=pi/2 +-npiPlease all I know is good explanation!!!

First, let's say 2x = y

So, for now, we are looking for places where tan y = 1

The unit circle is useful for figuring out the places in `[0,2 pi)`

where trig functions behave a certain way.  Then, we can use the fact that sine and cosine (and therefore tangent as well) have a period of `2 pi`

So:

On the unit circle, the x-value of a point represents the cosine and the y-value represents the sine of the angle made with the horizontal.

tangent = sine/cosine

Thus, if tangent = 1, we are looking for the places where sine x = cosine x.  In other words, we are looking for the places where the circle intersects the line y=x

This happens at `pi/4` and `(5 pi) / 4`

So:

`"tan" pi/4 = 1` and `"tan" (5 pi ) / 4 = 1`

Now, sine and cosine are periodic, with period `2 pi `

So, if `"tan" pi / 4 = 1`

then we also know that:

`..., "tan" (pi / 4 - 4 pi) = 1, "tan" (pi/4 - 2 pi) = 1, "tan" (pi/4) = 1, "tan" (pi/4 + 2 pi) = 1, ...`

So, some possibilities for y are:

`pi/4 - 4 pi`, `pi/4 - 2 pi`, `pi/4`, `pi/4 + 2 pi`, `pi/4 + 4 pi`, ...

in other words,

`pi/4 + 2k pi`, where k is some whole number

But, we also have that `"tan" (5 pi)/4 = 1`

Thus we also know that

`..., "tan" ((5 pi)/4 - 4 pi) = 1, "tan" ((5 pi)/4 - 2 pi) = 1, "tan" ((5 pi)/4 ) = 1, "tan" ((5 pi)/4 + 2 pi) = 1,... `

in other words, `(5 pi)/4 + 2k pi`, where k is some whole number

But, `(5 pi)/4 + 2k pi = pi/4 + pi + 2k pi`

In other words, we can write all of these, combined, as `pi/4 + k pi`, where k is some whole number

Thus, `y = pi/4 + k pi`, k a whole number

So  `2x = pi/4 + k pi`, k a whole number

So `x = pi/8 + (k/2) pi`, k a whole number

These are all the possible x which satisfy `"tan" 2x = 1`

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