Question

Решите уравнение sin2x-2√3 cos(x+7π/6)=3cosx [-3π/2;0]

Answer 1

2sin x*cos x - 2√3*(cos x*cos(7П/6) - sin x*sin(7П/6)) = 3cos x

2sin x*cos x - 2√3*(cos x*(-√3/2) - sin x*(-1/2)) = 3cos x

2sin x*cos x + 2√3*√3/2*cos x - 2√3/2*sin x = 3cos x

2√3*√3/2*cos x = 3cos x, их можно сократить.

2sin x*cos x - √3*sin x = 0

sin x*(2cos x - √3) = 0

1) sin x = 0; x = П*k, k € Z. На отрезке [-3П/2; 0] будут корни x1 = -П; x2 = 0

2) cos x = √3/2; x = +-П/6 + 2П*n, n € Z. На отрезке [-3П/2; 0] будет x3 = -П/6

Ответ: x1 = -П; x2 = 0; x3 = -П/6