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