Question

Solving a trigonometric equation involving an angle multiplied by a constant

Answer 1

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

$2k\pi,k\in\Z$

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:

$\begin{gathered} \sqrt[]{3}\tan (3x)+1=0 \\ \sqrt[]{3}\tan (3x)=-1 \\ \tan (3x)=-\frac{1}{\sqrt[]{3}} \\ \tan (3x)=-\frac{\sqrt[]{3}}{3} \end{gathered}$

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

$+\frac{\sqrt[]{3}}{3}$

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

$-\frac{\sqrt[]{3}}{3}$

Are:

$\begin{gathered} \theta_1=\pi-\frac{\pi}{6}=\frac{5\pi}{6} \\ \theta_2=2\pi-\frac{\pi}{6}=\frac{11\pi}{6} \end{gathered}$

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

$\begin{gathered} \theta=\frac{5\pi}{6}+2k\pi,k\in\Z \\ or \\ \theta=\frac{11\pi}{6}+2k\pi,k\in\Z \end{gathered}$

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

$\theta=\frac{5\pi}{6}+k\pi,k\in\Z$

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

$\begin{gathered} 3x=\theta \\ 3x=\frac{5\pi}{6}+k\pi,k\in\Z \\ x=\frac{5\pi}{18}+\frac{k\pi}{3},k\in\Z \end{gathered}$

So, the solutions are:

$x=\frac{5\pi}{18}+\frac{k\pi}{3},k\in\Z$