# Question #f7c16

Feb 8, 2018

$x = \frac{\left(1 + 4 k\right) \pi}{12}$ where $k \in Z$.

See explanation.

#### Explanation:

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

First of all, since I never remember my trigonometric functions, I will rewrite the equation as such:
$\tan 3 x = \frac{\sin 3 x}{\cos 3 x} = 1$
This can be done because $\tan \theta = \frac{\sin \theta}{\cos \theta}$ only when $\cos \theta$ is not equal to $0$ (otherwise the division does not work) and this means whenever $\theta = \frac{\pi}{2} + k \pi$ 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
$\sin 3 x = \cos 3 x$

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

In our case, $\theta = 3 x$, and so this last equation becomes:
$3 x = \frac{\pi}{4} + k \pi$
that is:
$x = \frac{1}{3} \left(\frac{\pi}{4} + k \pi\right)$
Simplifying a bit more gives:
$x = \frac{1}{3} \frac{\pi + 4 k \pi}{4}$
and finally
$x = \frac{\left(1 + 4 k\right) \pi}{12}$

Okay, this means that whenever $x$ is of that form, $\sin 3 x = \cos 3 x$, which means that $\tan 3 x = 1$.
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, $x = \frac{\left(1 + 4 \cdot 3\right) \pi}{12} = \frac{13 \pi}{12}$
Now, $3 x = 3 \cdot \frac{13 \pi}{12} = \frac{13 \pi}{4}$
That is, $\sin 3 x = \sin \left(\frac{13 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$
and, $\cos 3 x = \cos \left(\frac{13 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$
so they are equal,
and therefore, $\tan 3 x = 1$.