Question

solve for x: 3tan^3x-tanx=0

Answer 1

Answer on Question #62719 – Math – Trigonometry

Question

Solve for $x$:

$$3 \tan^3 x - \tan x = 0$$

Solution

$$3 \tan^3 x - \tan x = 0$$ $$\tan x \cdot (3 \tan^2 x - 1) = 0$$

The following cases are possible:

1) $\tan(x) = 0$

$$x = \pi k, \; k \in \mathbb{Z}$$

2) $3 \tan^2(x) - 1 = 0$

$$3 \tan^2(x) = 1$$ $$\tan^2(x) = \frac{1}{3} \Leftrightarrow \tan(x) = \frac{1}{\sqrt{3}} \quad \text{or} \quad \tan(x) = -\frac{1}{\sqrt{3}}$$

If $\tan(x) = \frac{1}{\sqrt{3}}$, then

$$\begin{aligned} x &= \arctan \left(\frac{1}{\sqrt{3}}\right) + l\pi, \; l \in \mathbb{Z} \\ x &= \frac{\pi}{6} + l\pi, \; l \in \mathbb{Z}. \end{aligned}$$

If $\tan(x) = -\frac{1}{\sqrt{3}}$, then

$$\begin{aligned} x &= \arctan \left(-\frac{1}{\sqrt{3}}\right) + m\pi, \; m \in \mathbb{Z} \\ x &= -\frac{\pi}{6} + m\pi, \; m \in \mathbb{Z}. \end{aligned}$$

Thus solutions of equation $3 \tan^3 x - \tan x = 0$ are $x = \pi k$, $x = \frac{\pi}{6} + l\pi$, $x = -\frac{\pi}{6} + m\pi$, $k \in \mathbb{Z}$, $l \in \mathbb{Z}$, $m \in \mathbb{Z}$.

**Answer**: $x = \pi k$, $x = \frac{\pi}{6} + l\pi$, $x = -\frac{\pi}{6} + m\pi$, $k \in \mathbb{Z}$, $l \in \mathbb{Z}$, $m \in \mathbb{Z}$.

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