In these questions, we need to follow the steps:
1 - solve for the trigonometric function
2 - Use the unit circle or a calculator to find which angles between 0 and 2π gives that results.
3 - Complete these angles with the complete round repetition, by adding

4 - these solutions are equal to the part inside the trigonometric function, so equalize the part inside with the expression and solve for <em>x</em> to get the solutions.
1 - To solve, we just use algebraic operations:
![\begin{gathered} \sqrt[]{3}\tan (3x)+1=0 \\ \sqrt[]{3}\tan (3x)=-1 \\ \tan (3x)=-\frac{1}{\sqrt[]{3}} \\ \tan (3x)=-\frac{\sqrt[]{3}}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Csqrt%5B%5D%7B3%7D%5Ctan%20%283x%29%2B1%3D0%20%5C%5C%20%5Csqrt%5B%5D%7B3%7D%5Ctan%20%283x%29%3D-1%20%5C%5C%20%5Ctan%20%283x%29%3D-%5Cfrac%7B1%7D%7B%5Csqrt%5B%5D%7B3%7D%7D%20%5C%5C%20%5Ctan%20%283x%29%3D-%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B3%7D%20%5Cend%7Bgathered%7D)
2 - From the unit circle, we can see that we will have one solution from the 2nd quadrant and one from the 4th quadrant:
The value for the angle that give positive
![+\frac{\sqrt[]{3}}{3}](https://tex.z-dn.net/?f=%2B%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B3%7D)
is known to be 30°, which is the same as π/6, so by symmetry, we can see that the angles that have a tangent of
![-\frac{\sqrt[]{3}}{3}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B3%7D)
Are:

3 - to consider all the solutions, we need to consider the possibility of more turn around the unit circle, so:

Since 5π/6 and 11π/6 are π radians apart, we can put them together into one expression:

4 - Now, we need to solve for <em>x</em>, because these solutions are for all the interior of the tangent function, so:

So, the solutions are: