Question

(cos3x+cos5x)+cos4x=0​

Answer 1

Ответ:

решение смотри на фотографии

Пошаговое объяснение:

Answer 2

To solve the trigonometric equation $\cos(3x) + \cos(5x) + \cos(4x) = 0$, you can use some trigonometric identities to simplify and find the values of $x$. Here's a step-by-step solution:

1. Start by using the sum-to-product trigonometric identity to rewrite $\cos(3x) + \cos(5x)$ as follows:

$\cos(3x) + \cos(5x) = 2\cos\left(\frac{3x + 5x}{2}\right)\cos\left(\frac{5x - 3x}{2}\right)$

Simplify this to:

$2\cos(4x)\cos(x)$

2. Now, your equation becomes:

$2\cos(4x)\cos(x) + \cos(4x) = 0$

3. Factor out the common factor $\cos(4x)$:

$\cos(4x)(2\cos(x) + 1) = 0$

4. Now, you have two factors to consider:

a) $\cos(4x) = 0$

To solve this equation, you can find the values of $x$ that make $\cos(4x) = 0$:

$4x = \frac{\pi}{2} + k\pi$ where $k$ is an integer.

Divide both sides by 4:

$x = \frac{\pi}{8} + \frac{k\pi}{4}$

b) $2\cos(x) + 1 = 0$

To solve this equation, isolate $\cos(x)$:

$2\cos(x) = -1$

$\cos(x) = -\frac{1}{2}$

Now, find the values of $x$ that satisfy $\cos(x) = -\frac{1}{2}$. This occurs in the second and third quadrants. In the unit circle, $\cos(\pi/3) = \frac{1}{2}$, so $\cos(\pi - \pi/3) = -\frac{1}{2}$. Therefore, you have two solutions in the interval [0, 2π):

$x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ (second quadrant)

and

$x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ (third quadrant)

So, the solutions to the original equation are:

$x = \frac{\pi}{8} + \frac{k\pi}{4}$ (for the $\cos(4x) = 0$ factor)

and

$x = \frac{2\pi}{3}$ and $x = \frac{5\pi}{3}$ (for the $\cos(x) = -\frac{1}{2}$ factor)