Question

Sin2xcos2x-cosxsinx=0

Answer 1

Ответ:

cos 2x cos x – sin 2x sin x =0

cos(2x+x)=0

cos3x=0

3x=П/2+Пn

x=П/6+Пn/3

Ответ:П/6+Пn/3

Пошаговое объяснение:

Answer 2

To solve the trigonometric equation $\sin(2x) \cos(2x) - \cos(x) \sin(x) = 0$, we can use trigonometric identities to simplify the equation and then solve for $x$. Let's break down the steps:

1. Apply the double angle identity for sine: $\sin(2x) = 2 \sin(x) \cos(x)$.
2. Apply the double angle identity for cosine: $\cos(2x) = \cos^2(x) - \sin^2(x)$.
3. Substitute the above identities into the equation:

$2 \sin(x) \cos(x) (\cos^2(x) - \sin^2(x)) - \cos(x) \sin(x) = 0$

4. Distribute the terms and simplify:

$2 \sin(x) \cos(x) \cos^2(x) - 2 \sin(x) \cos(x) \sin^2(x) - \cos(x) \sin(x) = 0$

5. Factor out $\sin(x) \cos(x)$:

$\sin(x) \cos(x) (2 \cos^2(x) - 2 \sin^2(x) - 1) = 0$

6. Use the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$:

$\sin(x) \cos(x) (2 \cos^2(x) - 2 (1 - \cos^2(x)) - 1) = 0$

7. Simplify further:

$\sin(x) \cos(x) (2 \cos^2(x) - 2 + 2 \cos^2(x) - 1) = 0$ $\sin(x) \cos(x) (4 \cos^2(x) - 3) = 0$

8. Now, the equation can be satisfied if either $\sin(x) = 0$ or $4 \cos^2(x) - 3 = 0$.

For $\sin(x) = 0$, this implies $x = k \pi$ where $k$ is an integer.

For $4 \cos^2(x) - 3 = 0$, solve for $\cos(x)$:

$4 \cos^2(x) = 3$ $\cos^2(x) = \frac{3}{4}$ $\cos(x) = \pm \frac{\sqrt{3}}{2}$

This gives us $x = \frac{\pi}{6} + 2\pi n$ or $x = \frac{11\pi}{6} + 2\pi n$, where $n$ is an integer.

In summary, the solutions to the equation $\sin(2x) \cos(2x) - \cos(x) \sin(x) = 0$ are $x = k \pi$, $x = \frac{\pi}{6} + 2\pi n$, and $x = \frac{11\pi}{6} + 2\pi n$, where $k$ and $n$ are integers.