Question

Решите уравнение: cos3x/sin2x=-sinx

Answer 1

cos(3x) = cos(x + 2x) = cosx*cos(2x) - sinx*sin(2x)
(cosx*cos(2x)/sin(2x)) - (sinx*sin(2x)/sin(2x)) = -sinx
cosx*(2cos^2(x) - 1)/(2sinx*cosx) - sinx = -sinx
(2cos^2(x) - 1)/(2sinx) = 0
2cos^2(x) - 1 = 0
cos^2(x) = 1/2
1) cosx = √2/2
x = +-π/4 + 2πk
2) cosx = -√2/2
x = +-3π/4 + 2πk
Объединяем решения 1) и 2), получаем:
x = π/4 + πk/2