Question

помогите в решении: sin 5x = sin x + sin 2x

Answer 1

sin(5x) = sin(x) + sin(2x) sin(5x) = sin(2x + 3x) = sin(2x)*cos(3x) + cos(2x)*sin(3x) = = 2sin(x)*cos(x)*cos(3x) + cos(2x)*(3sin(x) - 4sin^3(x)) = = sin(x)*[2cos(x)*cos(3x) + 3cos(2x) - 4cos(2x)*sin^2(x)] = = sin(x)*[2cos(x)*cos(3x) + 3cos(2x) - 4cos(2x)*(1 - cos(2x))/2] = = sin(x)*[2cos(x)*cos(3x) + 3cos(2x) - 2cos(2x) + 2cos^2(2x)] = = sin(x)*[2cos(x)*cos(3x) + cos(2x) + 2cos^2(2x)] Получаем sin(x)*[2cos(x)*cos(3x) + cos(2x) + 2cos^2(2x))] = sin(x) + 2sin(x)*cos(x) = sin(x)*(1 + 2cos(x)) 1) sin x = 0, x1 = pi*k 2) 2cos(x)*cos(3x) + cos(2x) + 2cos^2(2x) = 1 + 2cos(x) 2cos(x)*cos(3x) + cos(2x) + 2cos^2(2x) - 2cos(x) - 1 = 0 2cos(x)*cos(3x) + cos(2x) + 2cos^2(2x) - 1 - 2cos(x) = 0 2cos(x)*cos(3x) - 2cos(x) + cos(2x) + cos(4x) = 0 2cos(x)*(cos(3x) - 1) + 2cos(3x)*cos x = 0 2cos x*(cos(3x) - 1 + cos(3x)) = 0 cos x*(2cos(3x) - 1) = 0 3) cos x = 0, x2 = pi/2 + pi*k cos(3x) = 1/2, x3 = 1/3*(pi/3 + 2pi*k) = pi/9 + 2pi/3*k, x4 = 1/3*(-pi/3 + 2pi*k) = -pi/9 + 2pi/3*k Ответ: x1 = pi*k, x2 = pi/2 + pi*k, x3 = pi/9 + 2pi/3*k, x4 = -pi/9 + 2pi/3*k