cos (2π-x)+sin(π/2+x)=√2

спасибо

Cos(2Pi - x) + sin(Pi/2 + x) =√2=>
cos(-x) + cosx = √2=>
2cosx = (√2)=>
cosx = (√2)/2 => x = ±arccos((√2)/2) + 2Pi*n, n∈Z
x= ±Pi/4 + 2Pi*n, n∈Z=> так