Question

2 sin2(π2 − x) + sin 2x = 0

Answer 1

Ответ:

х=$\frac{k\pi}{2}$, где k€Z

Пошаговое объяснение:

sin(2(2π-x))=-sin2x

2 sin2(π2 − x) + sin 2x = 0

-2sin2x+sin2x=0

-sin2x=0

2x=πk

x=πk/2

Answer 2

To solve the equation $2 \sin^2\left(\frac{\pi}{2} - x\right) + \sin^2(2x) = 0$, we can use trigonometric identities and algebraic manipulation to simplify the equation and find its solutions.

Let's start by using the trigonometric identity $\sin^2(\theta) + \cos^2(\theta) = 1$ for any angle $\theta$. Applying this identity to the equation, we get:

$2 \sin^2\left(\frac{\pi}{2} - x\right) + \sin^2(2x) = 2 \cos^2(x) + \sin^2(2x) = 2(1 - \sin^2(x)) + \sin^2(2x) = 2 - 2\sin^2(x) + \sin^2(2x)$

Now, we can use the double angle formula for sine: $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. Applying this formula to the $\sin^2(2x)$ term:

$\sin^2(2x) = (2\sin(x)\cos(x))^2 = 4\sin^2(x)\cos^2(x)$

Substitute this back into the equation:

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

Now, we can factor out $2\sin^2(x)$:

$2\sin^2(x) (2\cos^2(x) - 1) = 0$

Since we're looking for solutions where the equation equals 0, we have two cases to consider:

1. $2\sin^2(x) = 0$ This gives us $\sin(x) = 0$, which means $x = k\pi$ for any integer $k$.

2. $2\cos^2(x) - 1 = 0$ Solving for $\cos^2(x)$: $\cos^2(x) = \frac{1}{2}$ $\cos(x) = \pm \frac{\sqrt{2}}{2}$

This means $x = \frac{\pi}{4} + 2k\pi$ or $x = \frac{3\pi}{4} + 2k\pi$ for any integer $k$.

In summary, the solutions to the equation $2 \sin^2\left(\frac{\pi}{2} - x\right) + \sin^2(2x) = 0$ are:

• $x = k\pi$ for any integer $k$,
• $x = \frac{\pi}{4} + 2k\pi$ for any integer $k$,
• $x = \frac{3\pi}{4} + 2k\pi$ for any integer $k$.