Question

Если 4sinx*cosx=√2, то х=?

Answer 1

$4sin(x)cos(x) = \sqrt{2}\\2sin(2x) = \sqrt{2}\\sin(2x) = \frac{1}{\sqrt{2}}\\2x = (-1)^n\frac{\pi}{4} + \pi n, n \in Z\\x = (-1)^n\frac{\pi}{8} + \frac{\pi}{2} n, n\in Z\\Answer: x = (-1)^n\frac{\pi}{8} + \frac{\pi}{2} n, n\in Z$

Answer 2

2*(2sinx*cosx)=2sin2x    (синус двойного угла)

2sin2x=√2

sin2x=√2/2

2x=π/4+2πn n∈Z; 2х=3π/4+2πn n∈Z

Ответ:

х=π/8+πn n∈Z;

x=3π/8+πn n∈Z.

Answer 3

To solve the equation 4sin(x)cos(x) = √2, we can use the trigonometric identity:

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

Applying this identity to the equation, we have:

2sin(2x) = √2.

Dividing both sides of the equation by 2, we get:

sin(2x) = √2 / 2.

Now, we need to find the angle whose sine is equal to √2 / 2. From the unit circle or trigonometric tables, we know that the angle whose sine is √2 / 2 is π/4 radians or 45 degrees.

Therefore, we have:

2x = π/4.

Dividing both sides of the equation by 2, we obtain:

x = π/8.

So, x is equal to π/8 or approximately 0.3927 radians.