sin^4x+cos2x=1
sin^4x+1-2sin^2x-sin^2x=1-sin^2x
sin^4x-2sin^2x=0
sin^2x(sin^2x-2)=0
1) sin^2x=0
sinx=0
x=pik, k∈Z
2) sin^2x=2
sinx=±√2
x=(-1)^k*arcsin(√2)+pik, k ∈Z
x=(-1)^(k+1)*arcsin(√2)+pik, k ∈Z
ОТВЕТ:
pik, k∈Z
(-1)^k*arcsin(√2)+pik, k ∈Z
(-1)^(k+1)*arcsin(√2)+pik, k ∈Z