Question

cos 3x*cos 2x-sin x*sin 6x=cos 7x

Answer 1

$cos(3x)cos(2x) - sin(x)sin(6x) = cos(7x)\cos(3x)cos(2x) - sin(x)sin(6x) = cos(x + 6x)\cos(3x)cos(2x) - sin(x)sin(6x) = cos(x)cos(6x) - sin(x)sin(6x)\cos(3x)cos(2x) = cos(x)cos(6x)\frac{1}{2}(cos(x) + cos(5x)) = frac{1}{2}(cos(x) + cos(7x))\cos(5x) = cos(7x)\5x = pm7x + 2pi k, k in mathbb{Z}\left { {{x = pi n, n in mathbb{Z}} atop {x=frac{pi m}{6}, m in mathbb{Z}}} right. \x = frac{pi m}{6}, m in mathbb{Z}$

Корни из второго уравнения системы включают в себя корни первой.