Solve the equation 3 tan x + sqrt(3) = 0, giving the general solution with k an integer.

Step-by-step solution

1. 1

   Isolate the tangent. Move the constant and divide by $3$:

   $\tan x = -\frac{\sqrt{3}}{3}$

   It helps to rationalise this into a familiar shape: $\dfrac{\sqrt3}{3} = \dfrac{1}{\sqrt3}$, so $\tan x = -\dfrac{1}{\sqrt3}$.

2. 2

   Find the reference angle. $\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt3}$, from the $30^{\circ}$–$60^{\circ}$–$90^{\circ}$ triangle with sides $1$, $\sqrt3$, $2$. So the reference angle is $\dfrac{\pi}{6}$, i.e. $30^{\circ}$ — not $\dfrac{\pi}{3}$, which is where $\tan = \sqrt3$.

3. 3

   Use the sign to place the angle. Tangent is negative in the second and fourth quadrants. Because $\tan$ is an odd function, the principal solution is simply the negative of the reference angle:

   $x = -\frac{\pi}{6}$

4. 4

   Add the period once, not twice. Tangent has period $\pi$, not $2\pi$, and every solution is captured by a single family:

   $x = -\frac{\pi}{6} + \pi k, \qquad k \in \mathbb{Z}$

   Writing $-\tfrac{\pi}{6} + 2\pi k$ would lose half the solutions; writing $\tfrac{\pi}{6} + \pi k$ would give $\tan x = +\tfrac{1}{\sqrt3}$ and the wrong sign.

5. 5

   Verify. At $k=0$: $\tan\!\left(-\tfrac{\pi}{6}\right) = -\tfrac{\sqrt3}{3}$, and $3\left(-\tfrac{\sqrt3}{3}\right) + \sqrt3 = -\sqrt3 + \sqrt3 = 0$ ✓. At $k=1$: $x = \tfrac{5\pi}{6}$, and $\tan\tfrac{5\pi}{6} = -\tfrac{\sqrt3}{3}$ ✓, a second-quadrant solution as expected.