Questions: Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.) 3 tan^3(θ) = tan(θ) θ =

To solve the equation \(3 \tan^3(\theta) = \tan(\theta)\), we can first factor the equation. This involves setting the equation to zero and factoring out the common term. After factoring, we solve for \(\tan(\theta)\) and find the corresponding \(\theta\) values. We also consider the periodic nature of the tangent function to express the general solution.

Comenzamos con la ecuación dada:

\[ 3 \tan^3(\theta) = \tan(\theta) \]

Reorganizamos la ecuación para igualarla a cero:

\[ 3 \tan^3(\theta) - \tan(\theta) = 0 \]

Factorizamos la ecuación:

\[ \tan(\theta)(3 \tan^2(\theta) - 1) = 0 \]

Esto nos da dos casos a considerar:

1. \(\tan(\theta) = 0\)
2. \(3 \tan^2(\theta) - 1 = 0\)

Para el primer caso, \(\tan(\theta) = 0\):

\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]

Para el segundo caso, resolvemos \(3 \tan^2(\theta) - 1 = 0\):

\[ \tan^2(\theta) = \frac{1}{3} \implies \tan(\theta) = \pm \frac{1}{\sqrt{3}} \]

Esto nos da:

\[ \theta = \frac{\pi}{6} + n\pi \quad \text{y} \quad \theta = -\frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

Las soluciones específicas que encontramos son:

1. \(\theta = 0\)
2. \(\theta = -\frac{\pi}{6}\)
3. \(\theta = \frac{\pi}{6}\)

Las soluciones son:

\[ \boxed{\theta = 0, -\frac{\pi}{6}, \frac{\pi}{6}} \]