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.