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
[tex]2k\pi,k\in\Z[/tex]
4 - these solutions are equal to the part inside the trigonometric function, so equalize the part inside with the expression and solve for x to get the solutions.
1 - To solve, we just use algebraic operations:
[tex]\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}[/tex]
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
[tex]+\frac{\sqrt[]{3}}{3}[/tex]
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
[tex]-\frac{\sqrt[]{3}}{3}[/tex]
Are:
[tex]\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}[/tex]
3 - to consider all the solutions, we need to consider the possibility of more turn around the unit circle, so:
[tex]\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}[/tex]
Since 5π/6 and 11π/6 are π radians apart, we can put them together into one expression:
[tex]\theta=\frac{5\pi}{6}+k\pi,k\in\Z[/tex]
4 - Now, we need to solve for x, because these solutions are for all the interior of the tangent function, so:
[tex]\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}[/tex]
So, the solutions are:
[tex]x=\frac{5\pi}{18}+\frac{k\pi}{3},k\in\Z[/tex]
