Question

1+sqrt(3)*ctg(2x)=0 {0,2pi}

Answer 1

$1 + \sqrt{3 } \cot(2x) = 0$
ОДЗ: ( здесь вместо равно должно быть перечеркнутый знак равно)
$\sin(2x) = 0 \\ 2x = \pi \: k \\ x = \frac{\pi \: k}{2}$
Здесь простой знак равенства.
$\sqrt{3} \cot(2x) = - 1 \\ \cot(2x) = - \frac{1}{ \sqrt{3} } \\ \cot(2x) = - \frac{ \sqrt{3} }{3} \\ 2x = \frac{2\pi}{3} + \pi \: k \: \\ x = \frac{\pi}{3} + \frac{\pi \: k}{2}$

Answer 2

To solve the equation $1 + \sqrt{3} \cdot \cot(2x) = 0$ over the interval $[0, 2\pi]$, we'll follow these steps:

Step 1: Rewrite the equation in terms of trigonometric functions. Step 2: Find the values of $x$ that satisfy the equation. Step 3: Check if the solutions lie within the given interval $[0, 2\pi]$.

Let's proceed with the solution:

Step 1: Rewrite the equation in terms of trigonometric functions.

The cotangent function ($\cot$) is the reciprocal of the tangent function ($\tan$). So, $\cot(2x) = \frac{1}{\tan(2x)}$. Using the identity $\tan(2x) = \frac{2 \tan(x)}{1 - \tan^2(x)}$, we can rewrite the equation as follows:

$1 + \sqrt{3} \cdot \frac{1}{\tan(2x)} = 0$

Step 2: Find the values of $x$ that satisfy the equation.

To solve for $x$, let's isolate the $\tan(2x)$ term:

$\sqrt{3} \cdot \frac{1}{\tan(2x)} = -1$

Now, flip the equation to get rid of the fraction:

$\tan(2x) = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

Step 3: Check if the solutions lie within the given interval $[0, 2\pi]$.

The solutions for $x$ can be found using the inverse tangent function:

$2x = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) + k\pi$

Where $k$ is an integer. Now, divide by 2 to get $x$:

$x = \frac{\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) + k\pi}{2}$

Since we are interested in solutions within the interval $[0, 2\pi]$, we need to find the values of $k$ that satisfy this condition.

Let's calculate the value of $\frac{\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)}{2}$ first:

$\frac{\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)}{2} \approx -0.5236$

In radians, $\pi \approx 3.14159$. To find the values of $k$ that lie within the interval $[0, 2\pi]$, we can set up the inequalities:

$0 \leq -0.5236 + k\pi \leq 2\pi$

Now, solve for $k$:

$0 \leq k\pi \leq 2\pi + 0.5236$

Divide by $\pi$ (since $\pi > 0$):

$0 \leq k \leq \frac{2\pi + 0.5236}{\pi}$

$0 \leq k \leq 2.5236$

Since $k$ must be an integer, the possible values for $k$ are 0, 1, and 2.

Now, calculate the corresponding values of $x$ for each $k$:

For $k = 0$:

$x = \frac{\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) + 0\pi}{2} \approx -0.5236$