Question

Синус 2 х+ косинус 2х=0

Answer 1

Cos4x - sin2x = 0
1 - 2sin(^2)2x - sin2x = 0
2sin^2)2x + sin2x - 1 = 0
D = 1 + 8 = 9
1)sin2x =( -1 + 3):4 = 1/2
2x = (-1)^n π/6 + πn, nZ
x = (-1)^n π/12 + πn/2, nZ
2) sin2x = -1
2x = -π/2 + 2πn, nZ
x = - π/4 + πn, nZ

Answer 2

To solve the equation $\sin(2x) + \cos(2x) = 0$, we'll use trigonometric identities to simplify it. First, we'll rewrite the trigonometric functions using double-angle identities:

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

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

Now, the equation becomes:

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

Next, let's combine the like terms:

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

Now, we'll use the Pythagorean identity $\cos^2(x) + \sin^2(x) = 1$ to replace $\cos^2(x) - \sin^2(x)$ with 1:

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

Now, we can solve for $\sin(x)\cos(x)$:

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

Now, we'll divide both sides by 2:

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

There are multiple solutions for $x$ that satisfy this equation. One example is $x = \frac{7\pi}{12}$. You can find other solutions using the unit circle or a calculator if you need more values.