Question

Cos 3x + cos 5x = 0; ​

Answer 1

Формула суммы косинусов:

$\cos\alpha +\cos\beta =2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha-\beta }{2}$

Рассмотрим уравнение:

$\cos3x+\cos5x=0$

$2\cos\dfrac{3x+5x}{2}\cos\dfrac{3x-5x}{2}=0$

$2\cos4x\cos(-x)=0$

$\cos4x\cos x=0$

Произведение равно нулю, когда один из множителей равен 0:

$\cos4x=0\Rightarrow 4x=\dfrac{\pi }{2} +\pi n\Rightarrow \boxed{x_1=\dfrac{\pi }{8} +\dfrac{\pi n}{4} .\ n\in\mathbb{Z}}$

$\cos x=0\Rightarrow\boxed{ x_2=\dfrac{\pi }{2} +\pi n .\ n\in\mathbb{Z}}$

Ответ: $\dfrac{\pi }{8} +\dfrac{\pi n}{4} ;\ \dfrac{\pi }{2} +\pi n .\ n\in\mathbb{Z}$

Answer 2

To solve the equation cos(3x) + cos(5x) = 0, we can use trigonometric identities and algebraic techniques. Let's break it down step by step:

Step 1: Use the sum-to-product trigonometric identity. cos(a) + cos(b) = 2 * cos((a + b)/2) * cos((a - b)/2)

Applying this identity, we have: 2 * cos((3x + 5x)/2) * cos((3x - 5x)/2) = 0

Simplifying further, we get: 2 * cos(4x) * cos(-x) = 0

Step 2: Use the zero product property. For the product of two factors to be zero, at least one of the factors must be zero.

Therefore, we have two cases to consider:

Case 1: cos(4x) = 0 To find the values of x that satisfy this condition, we solve: 4x = (2n + 1) * π/2, where n is an integer.

Dividing both sides by 4, we get: x = (2n + 1) * π/8, where n is an integer.

Case 2: cos(-x) = 0 This case occurs when -x is equal to an odd multiple of π/2, which means x is an even multiple of π/2.

Combining both cases, we have two families of solutions: x = (2n + 1) * π/8, where n is an integer, or x = m * π/2, where m is an even integer.

These are the solutions to the equation cos(3x) + cos(5x) = 0.