Question

Решить систему уравнений:

 

Sin ( x / 2 ) - Cos 2y = 1

2 Sin ^2 (x / 2) - 3 Cos 2y = 2

Answer 1

$sin ( x / 2 ) - cos 2y = 1\ 2 sin ^2 (x / 2) - 3 cos 2y = 2\ \ sin (x/2) = 1 + cos 2y\ 2 (1 + cos 2y)^2 - 3 cos 2y = 2\ 2 (1 + 2cos 2y + cos^2 2y) - 3 cos 2y = 2\ 2 + 4cos 2y + 2cos^2 2y - 3 cos 2y = 2\ cos 2y + 2cos^2 2y = 0\ cos 2y(1 + 2cos 2y) = 0\ cos 2y_1 = 0 ; cos 2y_2 = -1/2\ sin (x_1/2) = 1 ; sin (x_2/2) = 1/2\ \ y_1 = frac{pi}{4}+frac{pi k}{2}\ x_1 = pi+4pi k\ \ y_2 = pi k pm frac{2pi}{6}\ x_2 = pi + 4pi k pm frac{2pi}{3}$