Question

[tex]sin3x+sinx=0 gjvgub nyt gf;fkecnf

Answer 1

Ответ:

x=kn/2

Объяснение:

x=kn/2;

x=n/2+kn

Answer 2

It looks like you've entered a mathematical equation involving trigonometric functions. The equation is $\sin(3x) + \sin(x) = 0$. If you're looking to solve this equation for $x$, I can certainly help you with that.

To solve this equation, you can use the trigonometric sum-to-product identities. The sum-to-product identity states that $\sin(a) + \sin(b) = 2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)$. Applying this identity to your equation:

$\sin(3x) + \sin(x) = 2\sin\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right) = 2\sin\left(\frac{4x}{2}\right)\cos\left(\frac{2x}{2}\right) = 2\sin(2x)\cos(x)$.

Now, we have the equation $2\sin(2x)\cos(x) = 0$. This equation will be true if either $2\sin(2x) = 0$ or $\cos(x) = 0$.

1. $2\sin(2x) = 0$ implies that $\sin(2x) = 0$. This occurs when $2x = k\pi$, where $k$ is an integer. Solving for $x$, we get $x = \frac{k\pi}{2}$.

2. $\cos(x) = 0$ when $x = \frac{\pi}{2} + k\pi$, where $k$ is an integer.

So, the solutions for $x$ that satisfy the equation $\sin(3x) + \sin(x) = 0$ are $x = \frac{k\pi}{2}$ and $x = \frac{\pi}{2} + k\pi$, where $k$ can be any integer.

Please note that I've considered the solutions in the interval $0 \leq x < 2\pi$ since trigonometric functions are periodic. If you're looking for solutions in a different interval, you may need to adjust the range accordingly.