Question

2sin^2x+cos4x=0 (5п/2; 3п)

Answer 1

Формула косинуса двойного аргумента:
cos 2x = cos^2 x - sin^2 x = 2cos^2 x - 1 = 1 - 2sin^2 x
Отсюда 2sin^2 x = 1 - cos (2x). Подставляем
1 - cos (2x) + cos (4x) = 0
2cos^2 (2x) - 1 - cos (2x) + 1 = 0
2cos^2 (2x) - cos (2x) = 0
cos (2x)*(2cos (2x) - 1) = 0
1) cos (2x) = 0; 2x = pi/2 + pi*k; x = pi/4 + pi/2*k
В промежуток (5pi/2; 3pi) попадает корень x1 = pi/4 + 5pi/2 = 11pi/4
2) cos (2x) = 1/2; 2x = pi/3 + 2pi*n; x = pi/6 + pi*n
В промежуток (5pi/2; 3pi) не попадает ни один корень.
3) cos (2x) = 1/2; 2x = -pi/3 + 2pi*n; x = -pi/6 + pi*n
В промежуток (5pi/2; 3pi) попадает корень x2 = -pi/6 + 3pi = 17pi/6
Ответ: x1 = 11pi/4; x2 = 17pi/6

Answer 2

To solve the trigonometric equation $2\sin^2(x) + \cos(4x) = 0$ over the interval $(\frac{5\pi}{2}, 3\pi)$, we can follow these steps:

1. Simplify the equation by using double angle and trigonometric identities. Recall that $\cos(2\theta) = 1 - 2\sin^2(\theta)$, so we can rewrite $\cos(4x)$ as $\cos(2 \cdot 2x)$ and apply the double angle formula:

$$\cos(4x) = \cos(2 \cdot 2x) = 1 - 2\sin^2(2x)$$

So the equation becomes:

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

2. Now, let's solve for $\sin(2x)$:

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

3. Rearrange the terms:

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

4. Move all terms to one side of the equation:

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

5. Factor out $2\sin^2(x)$:

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

6. Apply the identity $\sin^2(2x) = 1 - \cos^2(2x)$:

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

7. Simplify further:

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

8. Divide both sides by -2:

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

9. Use the fact that $\sin^2(x) = 1 - \cos^2(x)$:

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

10. Now, we have two cases to consider:

Case 1: $\cos^2(x) = 0$

If $\cos^2(x) = 0$, then $\cos(x) = 0$. This happens when $x = \frac{\pi}{2} + n\pi$ for any integer $n$. However, since we are only interested in the interval $(\frac{5\pi}{2}, 3\pi)$, we need to find $n$ such that $\frac{5\pi}{2} < x < 3\pi$.

The solutions for this case are $x = \frac{5\pi}{2}$ and $x = \frac{7\pi}{2}$.

Case 2: $\cos^2(2x) = 0$

If $\cos^2(2x) = 0$, then $\cos(2x) = 0$. This happens when $2x = \frac{\pi}{2} + n\pi$ for any integer $n$. Solve for $x$:

$$2x = \frac{\pi}{2} + n\pi x = \frac{\pi}{4} + \frac{n\pi}{2}$$