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#.