Question

Sin2x+cos2x • sinx =0

Answer 1

Ответ:

$\sin(2x) + \cos(2x) \times \sin(x) = 0 \\ 2 \sin(x) \cos(x) + \sin(x) \cos(2x) = 0 \\ \sin(x) (2 \cos(x) + \cos(2x) ) = 0 \\ \\ \sin(x) = 0 \\ x_1 = \pi \: n \\ \\ 2 \cos(x) + \cos(2x) = 0 \\ 2 \cos(x) + 2 \cos {}^{2} (x) - 1 = 0 \\ 2 \cos {}^{2} (x) + 2 \cos(x) - 1 = 0 \\ \\ \cos(x) = t \\ \\ 2t {}^{2} + 2 t - 1 = 0\\ D = 4 + 12 = 16\\ t_1 = \frac{ - 2 + 4}{4} = \frac{1}{2} \\ t_2 = - \frac{6}{4} = - 1.5 \\ \\ \cos(x) = \frac{1}{2} \\ x_2 = \pm \frac{\pi}{3} + 2\pi \: n \\ \\ \cos(x) = - 1.5$

нет корней

Ответ:

$x_1 = \pi \: n \\ x_2 = \pm \frac{\pi}{3} + 2 \pi \: n$

n принадлежит Z

Answer 2

To solve the equation $\sin(2x) + \cos(2x) \cdot \sin(x) = 0$, you can use trigonometric identities to simplify the equation and then solve for $x$.

First, rewrite $\sin(2x)$ using the double angle formula:

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

Now the equation becomes:

$$2\sin(x)\cos(x) + \cos(2x)\sin(x) = 0.$$

Next, rewrite $\cos(2x)$ using the double angle formula:

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

Substitute this into the equation:

$$2\sin(x)\cos(x) + (\cos^2(x) - \sin^2(x))\sin(x) = 0.$$

Now distribute $\sin(x)$ and group like terms:

$$2\sin(x)\cos(x) + \cos^2(x)\sin(x) - \sin^3(x) = 0.$$

Factor out $\sin(x)$ from the last two terms:

$$\sin(x)(2\cos(x) + \cos^2(x) - \sin^2(x)) = 0.$$

Now, you can use trigonometric identities to further simplify the expression inside the parentheses. Remember the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$. Using this identity, you can simplify the expression inside the parentheses:

$$2\cos(x) + \cos^2(x) - \sin^2(x) = 2\cos(x) + \cos^2(x) - (1 - \cos^2(x)) = 3\cos(x) - 1.$$

So, the equation becomes:

$$\sin(x)(3\cos(x) - 1) = 0.$$

This equation is satisfied if either $\sin(x) = 0$ or $3\cos(x) - 1 = 0$.

1. Case 1: $\sin(x) = 0$

If $\sin(x) = 0$, this occurs when $x = k\pi$ where $k$ is an integer.

2. Case 2: $3\cos(x) - 1 = 0$

Solve $3\cos(x) - 1 = 0$ for $\cos(x)$:

$$3\cos(x) - 1 = 0 \implies \cos(x) = \frac{1}{3}.$$

Using the inverse cosine function, find $x$:

$$x = \cos^{-1}\left(\frac{1}{3}\right) \approx 1.23096 \text{ radians}.$$

So, the solutions to the given equation are $x = k\pi$ where $k$ is an integer, and $x \approx 1.23096$.