Question

Find all solutions in the interval [0, 2π).

7 tan3x - 21 tan x = 0

Help me please

Answer 1

Answer on Question #42396 – Math – Trigonometry

Find all solutions in the interval $[0, 2\pi)$:

$$7 \tan 3x - 21 \tan x = 0.$$

Solution.

$$\begin{aligned} \tan 3x &= \frac{\sin 3x}{\cos 3x} = \frac{3 \sin x - 4 \sin^3 x}{4 \cos^3 x - 3 \cos x} = \tan x \cdot \frac{3 - 4 \sin^2 x}{4 \cos^2 x - 3} = \tan x \cdot \frac{4 \cos^2 x - 1}{4 \cos^2 x - 3} = \\ &= \tan x \cdot \frac{\frac{4}{\tan^2 x + 1} - 1}{\frac{4}{\tan^2 x + 1} - 3} = \tan x \cdot \frac{3 - \tan^2 x}{1 - 3 \tan^2 x}; \end{aligned}$$

It follows $tg^2x + 1 = \frac{1}{\cos^2 x}$ from $\sin^2 x + \cos^2 x = 1$, therefore $\cos^2 x = \frac{1}{tg^2x + 1}$.

Solve

$$\begin{aligned} 7 \tan 3x - 21 \tan x &= 0 \Rightarrow \tan x \cdot \frac{3 - \tan^2 x}{1 - 3 \tan^2 x} - 3 \tan x = 0 \Rightarrow \left[ \tan x = 0, \frac{3 - \tan^2 x}{1 - 3 \tan^2 x} - 3 \right] = 0 \Rightarrow \\ &\Rightarrow \left[ \tan x = 0, 3 - \tan^2 x = 3 - 9 \tan^2 x \right] \Rightarrow \tan x = 0; \\ &\quad \left\{ \tan x = 0, x \in [0, 2\pi) \right. \Rightarrow \left[ x = 0, x = \pi \right]. \end{aligned}$$

Answer. $\{0; \pi\}$.

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