Question

4cos^2,5x=корень из cosx Укажите корни принадлежащие [5п/2;7п/2]

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

To find the roots of the equation $4\cos^2(5x) = \sqrt{\cos(x)}$ in the interval $\left[\frac{5\pi}{2}, \frac{7\pi}{2}\right]$, we'll first simplify the equation and then solve for $x$.

Step 1: Simplify the equation Start by noticing that $\cos^2(5x) = (\cos(5x))^2$ and $\sqrt{\cos(x)} = (\cos(x))^{1/2}$.

Now, rewrite the equation as follows: $4(\cos(5x))^2 = (\cos(x))^{1/2}$

Step 2: Convert everything to terms of $\cos(x)$ To simplify further, we'll use the identity $\cos(5x) = \cos^2(2x) - 1$.

The equation becomes: $4((\cos^2(2x) - 1))^2 = (\cos(x))^{1/2}$

Step 3: Substitute $\cos(2x)$ with $2\cos^2(x) - 1$ The identity $\cos(2x) = 2\cos^2(x) - 1$ will help us get rid of $\cos(2x)$.

The equation now becomes: $4((2\cos^2(x) - 1)^2 - 1)^2 = (\cos(x))^{1/2}$

Step 4: Simplify further Let $a = \cos(x)$. Now we have the equation: $4((2a - 1)^2 - 1)^2 = a^{1/2}$

Step 5: Solve for $a$ Let's solve for $a$: $4((2a - 1)^2 - 1)^2 = a^{1/2}$

$4((4a^2 - 4a + 1) - 1)^2 = a^{1/2}$

$4(4a^2 - 4a)^2 = a^{1/2}$

$16a^4 - 16a^3 = a^{1/2}$

Now, raise both sides to the power of 2 to get rid of the fractional exponent: $256a^8 - 512a^7 + 256a^6 = a$

$256a^8 - 512a^7 + 256a^6 - a = 0$

Step 6: Find the roots of the equation We have reduced the problem to solving the equation $256a^8 - 512a^7 + 256a^6 - a = 0$.

Unfortunately, this equation is a higher-order polynomial, and solving it exactly can be quite complex. To find the exact roots, we would need to use numerical methods or a computer algebra system.

However, since you mentioned that the roots need to be in the interval $\left[\frac{5\pi}{2}, \frac{7\pi}{2}\right]$, we can use numerical methods like the bisection method or Newton-Raphson method to approximate the roots within that interval. Alternatively, a computer algebra system can be used to find approximate solutions as well.

Please note that the solutions can be irrational numbers or involve complex numbers, so the approximation might be necessary.