Question

Cos 2x-cos x = cos 3x​

Answer 1

$\cos 2x-\cos x =\cos 3x$

$\cos 2x=\cos x +\cos 3x$

$\cos 2x=2\cos\dfrac{x+3x}{2}\cos\dfrac{x-3x}{2}$

$\cos 2x=2\cos2x\cos(-x)$

$\cos 2x=2\cos2x\cos x$

$\cos 2x-2\cos2x\cos x=0$

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

Уравнение распадается на два. Решаем первое:

$\cos2x=0$

$2x=\dfrac{\pi}{2} +\pi n$

$\boxed{x=\dfrac{\pi}{4} +\dfrac{\pi n}{2} ,\ n\in\mathbb{Z}}$

Решаем второе:

$1-2\cos x=0$

$2\cos x=1$

$\cos x=\dfrac{1}{2}$

$\boxed{x=\pm\dfrac{\pi}{3} +2\pi n,\ n\in\mathbb{Z}}$

Answer 2

To solve the trigonometric equation $\cos(2x) - \cos(x) = \cos(3x)$, we can use trigonometric identities and algebraic manipulation. Here's how to solve it step by step:

1. Start with the equation: $\cos(2x) - \cos(x) = \cos(3x)$

2. Use the trigonometric identity for the double angle: $\cos(2x) = 2\cos^2(x) - 1$

3. Substitute this identity into the equation: $2\cos^2(x) - 1 - \cos(x) = \cos(3x)$

4. Use the sum-to-product identity for cosine (cosine of a sum of angles): $\cos(3x) = \cos(2x + x) = \cos(2x)\cos(x) - \sin(2x)\sin(x)$

5. Substitute the sum-to-product identity into the equation: $2\cos^2(x) - 1 - \cos(x) = \cos(2x)\cos(x) - \sin(2x)\sin(x)$

6. Rearrange the equation and simplify: $2\cos^2(x) - \cos(2x)\cos(x) + \sin(2x)\sin(x) - 1 - \cos(x) = 0$

7. Combine the terms involving cosine using the double angle identity: $2\cos^2(x) - 2\cos^2(x) + 1 - \cos(x) = 0$

8. Simplify further: $-\cos(x) + 1 = 0$

9. Solve for $\cos(x)$: $\cos(x) = 1$

10. Since $\cos(x) = 1$ when $x = 2n\pi$ (where $n$ is an integer), the solutions for $x$ are $x = 2n\pi$, where $n$ is any integer.

So, the solutions for the equation $\cos(2x) - \cos(x) = \cos(3x)$ are $x = 2n\pi$, where $n$ is an integer.