Question

Sin5x*sin4x + cos6x*cos3x=0​

Answer 1

$sin5x\cdot sin4x+cos6x\cdot cos3x=0\\\frac{1}{2}\left ( cosx-cos9x \right )+\frac{1}{2}\left ( cos3x+cos9x \right )=0\\cosx+cos3x=0\Leftrightarrow 2cos2x\cdot cosx=0\\cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{\pi k}{2},k\in \mathbb{Z}\\cosx=0\Rightarrow x=\frac{\pi}{2}+\pi k,k\in \mathbb{Z}$

Answer 2

To solve the equation $\sin(5x) \cdot \sin(4x) + \cos(6x) \cdot \cos(3x) = 0$, we'll use trigonometric identities and algebraic manipulation to simplify the expression and solve for $x$.

First, let's use the product-to-sum identities for cosine and sine:

$$\sin(a) \cdot \sin(b) = \frac{1}{2}[\cos(a - b) - \cos(a + b)]$$

$$\cos(a) \cdot \cos(b) = \frac{1}{2}[\cos(a - b) + \cos(a + b)]$$

Applying these identities to the given equation, we get:

$$\frac{1}{2}[\cos(5x - 4x) - \cos(5x + 4x)] + \frac{1}{2}[\cos(6x - 3x) + \cos(6x + 3x)] = 0$$

Simplifying further:

$$\frac{1}{2}[\cos(x) - \cos(9x)] + \frac{1}{2}[\cos(3x) + \cos(9x)] = 0$$

Now, combine the terms with $\cos(x)$ and $\cos(3x)$:

$$\frac{1}{2}[\cos(x) + \cos(3x)] = 0$$

Multiply both sides by 2 to get rid of the fraction:

$$\cos(x) + \cos(3x) = 0$$

Now, let's solve this equation for $x$:

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

Using the cosine addition formula $\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)$, we get:

$$\cos(3x) = \cos(x)\cos(2x) - \sin(x)\sin(2x)$$

Substitute this into the equation:

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

Rearrange and factor out $\cos(x)$:

$$\cos(x)(\cos(2x) + 1) = \sin(x)\sin(2x)$$

Divide both sides by $\cos(x)$ (assuming $\cos(x) \neq 0$):

$$\cos(2x) + 1 = \tan(x)\sin(2x)$$

Now, we can substitute $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and simplify further:

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

Multiply both sides by $\cos(x)$ to eliminate the fraction:

$$\cos(2x)\cos(x) + \cos(x) = \sin(x)\sin(2x)$$

Simplify further:

$$\cos(3x) + \cos(x) = \sin(x)\sin(2x)$$

We've now transformed the original equation into a new form. Unfortunately, solving this equation for $x$ analytically may not be straightforward, and numerical methods may be needed to approximate the solutions.