Question

Arcosx=arctgx найти корень уравнения

Answer 1

Ответ:

$\sqrt{\frac{\sqrt{5}-1}{2}}$

Пошаговое объяснение:

$acrrosx=arctgx\\\\tg(arccosx)=tg(arctgx)\\\\\frac{\sqrt{1-x^2}}{x}=x,\; \; \; x\neq 0\\\\\sqrt{1-x^2}=x^2\\\\(\sqrt{1-x^2})^2=(x^2)^2\\\\1-x^2=x^4\\\\x^4+x^2-1=0\\\\y=x^2,\; \; \; x\neq 0=>y>0\\\\y^2+y-1=0\\\\D=1^2-4*1*(-1)=1+4=5\\\\y_1=\frac{-1+\sqrt{5}}{2}=\frac{\sqrt{5}-1}{2} \; (>0)\; \; \; \; y_2=\frac{-1-\sqrt{5}}{2}\; (x=\sqrt{y}\\\\x=\sqrt{\frac{\sqrt{5}-1}{2}}$

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$tg(arctgx)=x,\; x\in R\\\\tgx=\frac{\sqrt{1-cos^2x}}{cosx},\; x\in[0;\pi/2)\cup (\pi/2;\pi]\; => \\\\=>tg(arccosx)=\frac{\sqrt{1-cos^2(arccosx)}}{cos(arccosx)}= \frac{\sqrt{1-x^2}}{x},\; x\in(-1;0)\cup (0;1)$

Answer 2

Solving the Equation arccos(x) = arctan(x) to Find the Root

To solve the equation arccos(x) = arctan(x) and find the root, we can use numerical methods such as the bisection method or Newton's method. These methods allow us to approximate the value of x that satisfies the equation.

Let's use the bisection method to find the root of the equation.

1. Bisection Method: - The bisection method is an iterative numerical method that finds the root of a function within a given interval. - The method works by repeatedly dividing the interval in half and selecting the subinterval where the function changes sign. - By iteratively narrowing down the interval, we can approximate the root of the equation.

2. Applying the Bisection Method: - To apply the bisection method, we need to choose an initial interval [a, b] where the function changes sign. - We can start with an interval that contains the root of the equation. - Let's choose an initial interval [a, b] = [-1, 1] since the range of both arccos(x) and arctan(x) is [-1, 1].

3. Iterative Steps: - We will perform several iterations to narrow down the interval and approximate the root. - In each iteration, we will calculate the midpoint of the interval and evaluate the function at that point. - If the function value is close to zero, we consider it as the root. - Otherwise, we update the interval by replacing either a or b with the midpoint, depending on the sign of the function value.

4. Example Iteration: - Let's perform one iteration of the bisection method to illustrate the process. - In the first iteration, we calculate the midpoint of the interval [a, b] = [-1, 1] as x = (a + b) / 2 = 0. - We evaluate the function at x and check if it is close to zero. - If the function value is close to zero, we consider x as the root. - Otherwise, we update the interval based on the sign of the function value.

5. Repeat Iterations: - We repeat the iterations until we reach a desired level of accuracy or until we find a root that satisfies the equation.

Please note that the bisection method is just one of the numerical methods that can be used to solve equations. There are other methods like Newton's method, secant method, etc., which may provide faster convergence in some cases.

I hope this explanation helps! Let me know if you have any further questions.