Question

2*cos(x) + 2*sin(2x) = 0​

Answer 1

x = π/2 + π n ,   7 π /6 + 2 π n ,   11 π/ 6 + 2 π n    для всех целых  n

Answer 2

To solve the equation $2\cos(x) + 2\sin(2x) = 0$, we can use trigonometric identities and algebraic techniques. First, let's rewrite $\sin(2x)$ using the double-angle identity for sine:

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

Now, substitute this back into the original equation:

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

Simplify the equation:

$2\cos(x) + 4\sin(x)\cos(x) = 0$

Factor out the common term of $2\cos(x)$:

$2\cos(x)(1 + 2\sin(x)) = 0$

Now, we have two possibilities:

1. $2\cos(x) = 0$

This gives us $\cos(x) = 0$, which occurs at $\frac{\pi}{2} + k\pi$ and $\frac{3\pi}{2} + k\pi$, where $k$ is an integer.

2. $1 + 2\sin(x) = 0$

Solving for $\sin(x)$, we get $\sin(x) = -\frac{1}{2}$, which occurs in the third and fourth quadrants. The solutions for this part are $x = \frac{7\pi}{6} + 2k\pi$ and $x = \frac{11\pi}{6} + 2k\pi$, where $k$ is an integer.

So, the solutions to the original equation $2\cos(x) + 2\sin(2x) = 0$ are:

$x = \frac{\pi}{2} + k\pi, \quad \frac{3\pi}{2} + k\pi, \quad \frac{7\pi}{6} + 2k\pi, \quad \text{and} \quad \frac{11\pi}{6} + 2k\pi,$

where $k$ is an integer.