Question

Ctg(x)-3tgx=0 решите)

Answer 1

Ctg x - 3tg x = 0
ctg x² = 3
ctg x = √3 или ctg x = -√3
1) x = π/6 + πk, k принадлежит Z
2) x = π/6 + πk, k принадлежит Z

Ответ: x = π/6 + πk, k принадлежит Z

Answer 2

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Answer 3

To solve the equation $\cot(x) - 3\tan(x) = 0$, we need to find the values of $x$ that satisfy the equation.

First, let's rewrite the equation in terms of trigonometric functions using their definitions:

$\cot(x) = \frac{1}{\tan(x)}$

Now, we can substitute this into the equation:

$\frac{1}{\tan(x)} - 3\tan(x) = 0$

Next, let's get rid of the fraction by multiplying the entire equation by $\tan(x)$:

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

Now, let's rearrange the equation:

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

This is a quadratic equation in terms of $\tan(x)$. To solve for $\tan(x)$, we can use the quadratic formula:

$\tan(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 3$, $b = 0$, and $c = -1$. Plugging in these values:

$\tan(x) = \frac{\pm \sqrt{0^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}$ $\tan(x) = \frac{\pm \sqrt{12}}{6}$ $\tan(x) = \frac{\pm 2\sqrt{3}}{6}$ $\tan(x) = \pm \frac{\sqrt{3}}{3}$

Now, let's recall that the tangent function is positive in the first and third quadrants, and negative in the second and fourth quadrants. So, we have two cases to consider:

Case 1: $\tan(x) = \frac{\sqrt{3}}{3}$

To find the solutions for this case, we can use the inverse tangent function:

$x = \arctan\left(\frac{\sqrt{3}}{3}\right)$ $x \approx 30^\circ$

Case 2: $\tan(x) = -\frac{\sqrt{3}}{3}$

For this case:

$x = \arctan\left(-\frac{\sqrt{3}}{3}\right)$ $x \approx -30^\circ$

However, remember that the tangent function has a period of $180^\circ$ (or $\pi$ radians), which means it repeats every $180^\circ$. So, the general solution for $x$ would be:

$x = 30^\circ + 180^\circ n$ $x = -30^\circ + 180^\circ n$

where $n$ is an integer (positive, negative, or zero) representing the number of complete cycles. It accounts for all possible solutions within the desired range.

Therefore, the general solution to the equation $\cot(x) - 3\tan(x) = 0$ is:

$x = 30^\circ + 180^\circ n$ $x = -30^\circ + 180^\circ n$

where $n$ is an integer.