Ответ:

x=π/3+kπ; y=±π/4+2nπ

Объяснение:

1 + 2cos2x = 0

cos2x=-0,5

2x=±arccos(-0,5)+2kπ=±(π-arccos0,5)+2kπ=±2π/3+2kπ

x=±π/3+kπ

√6cosy - 4sinx = 2√3(1 + sin²y)

sin²y=1-cos²y

√6cosy - 4sinx = 2√3(1 + 1-cos²y)=2√3(2-cos²y)

2√3cos²y+√6cosy - 4sinx-4√3=0

2√3cos²y+√6cosy - 4sin(±π/3+kπ)-4√3=0

2√3cos²y+√6cosy - 4·(±√3/2)-4√3=0

2√3cos²y+√6cosy ± 2√3-4√3=0

2cos²y+√2cosy ± 2-4=0

1) x=-π/3+kπ

2cos²y+√2cosy - 2-4=0⇒2cos²y+√2cosy=6

2cos²y+√2cosy≤2+√2<6

2) x=π/3+kπ

2cos²y+√2cosy + 2-4=0

2cos²y+√2cosy-2=0

√2cos²y+cosy-√2=0

cosy=t, |t|≤1

√2t²+t-√2=0

D=1+8=9

t₁=(-1-3)/(2√2)=-√2<1

t₂=(-1+3)/(2√2)=√2/2

cosy=√2/2

y=±π/4+2nπ