Question

Arcctg(x+3)+arcctg(4x)=p/2​

Answer 1

Ответ:

$\frac{-3+\sqrt{10} }{2}$

Объяснение:

$arcctg(x+3)+arcctg(4x)=\frac{\pi}{2} \\ \\ ctg(arcctg(x+3)+arcctg(4x))=ctg\frac{\pi}{2} \\ \\ \frac{ctg(arcctg(x+3))*ctg(arcctg(4x))-1}{ctg(arcctg(x+3))+ctg(arcctg(4x))} =0 \\ \\ \frac{(x+3)*4x-1}{x+3+4x}=0 \\ \\ \frac{4x^2+12x-1}{5x+3}=0$

$4x^2+12x-1=0 \\ D=12^2-4*4*(-1)=144+16=160\\ \sqrt{D}=\sqrt{160}=\sqrt{16*10}=4\sqrt{10} \\ \\ x_{1,2}=\frac{-12\pm 4\sqrt{10} }{2*4}=\frac{-3\pm \sqrt{10}}{2}$

при прямой подстановке в исходное уравнение, выясняем, что х=(-3-√10)/2 - нам не подходит

Answer 2

To solve the equation $\text{arcctg}(x+3) + \text{arcctg}(4x) = \frac{p}{2}$, where $\text{arcctg}$ represents the inverse cotangent function, you can use trigonometric identities and properties. First, we'll rewrite this equation using the cotangent function:

$$\cot(\text{arcctg}(x+3)) + \cot(\text{arcctg}(4x)) = \frac{1}{\tan(\frac{p}{2})}$$

Now, we'll use the properties of inverse trigonometric functions.

1. $\text{arcctg}(\cot(\theta)) = \theta$, where $0 < \theta < \pi$.

2. $\cot(\text{arcctg}(y)) = y$, where $y$ is any real number.

Using these properties, we can simplify the equation:

$$(x+3) + (4x) = \frac{1}{\tan(\frac{p}{2})}$$

Simplify the left side:

$$5x + 3 = \frac{1}{\tan(\frac{p}{2})}$$

Now, we can isolate $x$:

$$5x = \frac{1}{\tan(\frac{p}{2})} - 3$$

$$x = \frac{1}{5\tan(\frac{p}{2})} - \frac{3}{5}$$

So, the solution for $x$ in terms of $p$ is:

$$x = \frac{1}{5\tan(\frac{p}{2})} - \frac{3}{5}$$

Please note that $\tan(\frac{p}{2})$ should not equal zero, as the cotangent function is undefined at those points.