Question

Cos7x cos6x=cos5x cos8x

Answer 1

произведение косинусов:

cos(a)*cos(b) ≡ (1/2)*( cos(a+b) + cos(a-b) ),

тогда

cos(7x)*cos(6x) ≡ (1/2)*( cos(7x+6x) + cos(7x-6x) ) ≡ (1/2)*( cos(13x) + cos(x) ),

cos(5x)*cos(8x) ≡ (1/2)*( cos(5x+8x) + cos(5x-8x) ) ≡ (1/2)*( cos(13x) + cos(3x) ),

Исходное уравнение равносильно следующему

(1/2)*( cos(13x) + cos(x) ) = (1/2)*( cos(13x) + cos(3x) ),

cos(x) = cos(3x),

cos(3x) - cos(x) = 0,

формула разности косинусов:

cos(a) - cos(b) ≡ -2*sin( (a-b)/2 )*sin( (a+b)/2),

поэтому cos(3x) - cos(x) ≡ -2*sin( (3x-x)/2)*sin( (3x+x)/2)≡ -2*sin(x)*sin(2x),

поэтому получаем следующее уравнение

-2*sin(x)*sin(2x) = 0,

sin(x)*sin(2x) = 0,

1) sin(x) = 0 или 2) sin(2x) = 0,

1) x=πm, m∈Z,

2) 2x = πn, n∈Z,

x = πn/2.

Видно, что множество решений 1) есть подмножество решений 2) (при четных n).

Ответ. x=πn, n∈Z.

Answer 2

To solve the equation $\cos(7x) \cdot \cos(6x) = \cos(5x) \cdot \cos(8x)$, you can use trigonometric identities to simplify it. One helpful identity is the product-to-sum identity:

$$\cos(A) \cdot \cos(B) = \frac{1}{2} \left[ \cos(A + B) + \cos(A - B) \right]$$

Applying this identity to your equation:

$$\frac{1}{2} \left[ \cos(7x + 6x) + \cos(7x - 6x) \right] = \frac{1}{2} \left[ \cos(5x + 8x) + \cos(8x - 5x) \right]$$

Simplify further:

$$\frac{1}{2} \left[ \cos(13x) + \cos(x) \right] = \frac{1}{2} \left[ \cos(13x) + \cos(3x) \right]$$

Now, you have:

$$\cos(13x) + \cos(x) = \cos(13x) + \cos(3x)$$

To solve for $x$, you can cancel out the common term $\cos(13x)$ from both sides of the equation:

$$\cos(x) = \cos(3x)$$

Now, you need to find the values of $x$ that satisfy this equation. To do that, you can use the angle addition identity:

$$\cos(A) = \cos(B) \Rightarrow A = \pm B + 2\pi n \quad \text{or} \quad A = \pm (2\pi - B) + 2\pi n$$

In this case, $A = x$ and $B = 3x$, so:

$$x = \pm 3x + 2\pi n \quad \text{or} \quad x = \pm (2\pi - 3x) + 2\pi n$$

Now, solve for $x$ in both equations:

1. $x = \pm 3x + 2\pi n$:

   $$x - 3x = 2\pi n \implies -2x = 2\pi n \implies x = -\pi n$$

2. $x = \pm (2\pi - 3x) + 2\pi n$:

   $$x = 2\pi - 3x + 2\pi n$$

Now, you have two equations for $x$:

1. $x = -\pi n$
2. $x = 2\pi - 3x + 2\pi n$

Let's solve the second equation:

$$x = 2\pi - 3x + 2\pi n$$

Add $3x$ to both sides:

$$4x = 2\pi - 2\pi n$$

Divide by 4:

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

Now, you have two sets of solutions:

1. $x = -\pi n$, where $n$ is an integer.
2. $x = \frac{1}{2}(\pi - \pi n)$, where $n$ is an integer.

These are the solutions to the equation $\cos(7x) \cdot \cos(6x) = \cos(5x) \cdot \cos(8x)$.