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
![2k\pi,k\in\Z](https://tex.z-dn.net/?f=2k%5Cpi%2Ck%5Cin%5CZ)
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:
![\begin{gathered} \theta_1=\pi-\frac{\pi}{6}=\frac{5\pi}{6} \\ \theta_2=2\pi-\frac{\pi}{6}=\frac{11\pi}{6} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctheta_1%3D%5Cpi-%5Cfrac%7B%5Cpi%7D%7B6%7D%3D%5Cfrac%7B5%5Cpi%7D%7B6%7D%20%5C%5C%20%5Ctheta_2%3D2%5Cpi-%5Cfrac%7B%5Cpi%7D%7B6%7D%3D%5Cfrac%7B11%5Cpi%7D%7B6%7D%20%5Cend%7Bgathered%7D)
3 - to consider all the solutions, we need to consider the possibility of more turn around the unit circle, so:
![\begin{gathered} \theta=\frac{5\pi}{6}+2k\pi,k\in\Z \\ or \\ \theta=\frac{11\pi}{6}+2k\pi,k\in\Z \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctheta%3D%5Cfrac%7B5%5Cpi%7D%7B6%7D%2B2k%5Cpi%2Ck%5Cin%5CZ%20%5C%5C%20or%20%5C%5C%20%5Ctheta%3D%5Cfrac%7B11%5Cpi%7D%7B6%7D%2B2k%5Cpi%2Ck%5Cin%5CZ%20%5Cend%7Bgathered%7D)
Since 5π/6 and 11π/6 are π radians apart, we can put them together into one expression:
![\theta=\frac{5\pi}{6}+k\pi,k\in\Z](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B5%5Cpi%7D%7B6%7D%2Bk%5Cpi%2Ck%5Cin%5CZ)
4 - Now, we need to solve for <em>x</em>, because these solutions are for all the interior of the tangent function, so:
![\begin{gathered} 3x=\theta \\ 3x=\frac{5\pi}{6}+k\pi,k\in\Z \\ x=\frac{5\pi}{18}+\frac{k\pi}{3},k\in\Z \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%203x%3D%5Ctheta%20%5C%5C%203x%3D%5Cfrac%7B5%5Cpi%7D%7B6%7D%2Bk%5Cpi%2Ck%5Cin%5CZ%20%5C%5C%20x%3D%5Cfrac%7B5%5Cpi%7D%7B18%7D%2B%5Cfrac%7Bk%5Cpi%7D%7B3%7D%2Ck%5Cin%5CZ%20%5Cend%7Bgathered%7D)
So, the solutions are: