Question

3sin^2 (x-3п/2)-2cos(3п/2+х)cos(п+х)+2sin^2(х-п)=2

Answer 1

Формулы приведения
sin(x-3pi/2) = sin(x-3pi/2+2pi) = sin(x+pi/2) = cos x
cos(x+3pi/2) = cos(x+3pi/2-2pi) = cos(x-pi/2) = cos(pi/2-x) = sin x
cos(pi+x) = -cos x
sin(x-pi) = sin(x-pi+2pi) = sin(x+pi) = -sin x
Подставляем
3cos^2 x - 2sin x*(-cos x) + 2(-sin x)^2 = 2sin^2 x + 2cos^2 x
Приводим подобные
cos^2 x + 2sin x*cos x = 0
cos x*(cos x + 2sin x) = 0
1) cos x = 0; x1 = pi/2 + pi*k
2) 2sin x = -cos x
tg x = -1/2; x2 = -arctg(1/2) + pi*n