Решите уравнение: sinx sin3x+cos4x=0

(sinx+sin5x) + sin3x=02sin[ (x+5x) / 2] * cos[ (x-5x) / 2] + sin3x=02sin (6x/2) * cos (-4x/2) + sin3x=02sin3x * cos2x + sin3x=0sin3x * (2cos2x+1) = 0sin3x=0 2cos2x+1=03x=2pi*k, k∈ (-∞; +∞) 2cos2x=-1 x=2/3 pi*k, k∈ (-∞; +∞) cos2x=-1/2 2x=pi - arccos (1/2) 2x = pi - pi/3 + pi*k, k∈ (-∞; +∞) 2x = 2pi/3 + pi*k, k∈ (-∞; +∞) x = pi/3 + pi/2 * k, k∈ (-∞; +∞)