Question

Y=sin√x+tg^2 2x решите

Answer 1

1) sin(2x-п/6)=+-√(3/4) = +-√3/2
sin(2x-П/6) = (-1)^k(+-П/3)+Пk
x= (-1)^k(+-П/6)+П/12+Пk/2
2) sinx =(sin^2x/(1+cosx)^2)*(1+cosx)
sinx = sin^2x/(1+cosx)
sinx(1-sinx/(1+cosx))=0
sinx = 0 --> x = Пn
1+cosx-sinx = 0
sinx-cosx = 1 (^2)
sin^2x-2sinxcosx+cos^2x=1
sin2x=0
2x = Пn
x = Пn/2
Ответ: Пn/2, n принадл. z