Решить уравнение

sinx+cosx=1-sin2x

Sinx+cosx=1-sin2x

sinx+cosx=t

(sinx+cosx) ²=t²

sin²x+2sinxcosx+cos²x=t²

2sinxcosx=t²-1

sin2x=t²-1

t=1 - (t²-1)

t=1-t²+1

t²+t-2=0

D=1+8=9=3²

t = (-1±3) / 2

t1=-2; t2=1

1) sinx+cosx=-2; x€∅

2) sinx+cosx=1

√2sin (π/4+x) = 1

sin (π/4+x) = √2/2

π/4+x = (-1) ⁿπ/4+πn

x=-π/4 + (-1) ⁿπ/4+πn

n=2k

x=2πk

n=2k+1

x=-π/2+π (2k+1) =

π/2+2πk