Question

Question #f7c16

Answer 1

`x = ((1+4k)pi)/12` where `k in Z`.

See explanation.

Explanation:

I assume you meant "what is the general solution `x` that satisfies the equation `tan 3x = 1`".

First of all, since I never remember my trigonometric functions, I will rewrite the equation as such:
`tan 3x = (sin 3x)/(cos 3x)=1`
This can be done because `tan theta = (sin theta)/(cos theta)` only when `cos theta` is not equal to `0` (otherwise the division does not work) and this means whenever `theta = pi/2 + k pi` where `k` can be any relative integer (...,-2, -1, 0, 1, 2, etc...).

Now we see that the equality is only true if and only if
`sin 3x = cos 3x`

Remembering the trigonometric circle, sine and cosine are only equal whenever the angle is 45 degree, or `pi/4` radian, and every other 180 degrees from there, that is 225 degree for the other main angle, or in radian `(5pi)/4`.
If we were to write an equation for these, it would be
`theta = pi/4 + k pi` where `k` is any relative integer `Z`.

In our case, `theta = 3x`, and so this last equation becomes:
`3x = pi/4 + k pi`
that is:
`x = 1/3 (pi/4 + k pi)`
Simplifying a bit more gives:
`x = 1/3 (pi + 4k pi)/4`
and finally
`x = ((1+4k)pi)/12`

Okay, this means that whenever `x` is of that form, `sin 3x = cos 3x`, which means that `tan 3x = 1`.
If you are skeptic of the result, you can try a few examples to see if it actually works!
E.g.: Let's say, for example, that `k=3`.
Then, `x = ((1+4*3)pi)/12 = (13 pi)/12`
Now, `3x = 3* (13 pi)/12 = (13 pi)/4`
That is, `sin 3x = sin ( (13 pi)/4) = -sqrt(2)/2`
and, `cos 3x = cos ( (13 pi)/4) = -sqrt(2)/2`
so they are equal,
and therefore, `tan 3x = 1`.