2 / tg^2 (x+5p) + 1 / sin (x-5p) - 4 = 0 Укажите корни принадлежащие отрезку [-p/2; p/2]

tg^2 (x+5p) = tg^2 x

sin (x-5p) = - sin x

2 / tg^2 (x) = 2c tg^2 (x)

2ctg^2 (x) = 2/sin^2 (x) - 2

2/sin^2 (x) - 1/sin (x) - 6=0

1/sin (x) = t

2t^2-t-6=0

t=2, t=-1,5

x=pi/6+2pi*n x=5pi/6+2pi*n

x=arcsin (-2/3) + 2pi*n x=pi-arcsin (-2/3) + 2pi*n

x=pi/6, arcsin (-2/3)