sin(x2)=1–2sin2x
sin(x2)=cos2x
sin(x2)=sin(π/2)–2x)
x2=(π/2)–2x+2πn, n ∈ Z
x2+2x–(π/2)–2πn=0, n ∈ Z
D=4–4·(–(π/2)–2πn)=4+2π+8πn, n ∈ Z
D ≥ 0
4+2π+8πn ≥ 0 ⇒ n ∈ Z n ≥ 0 ⇒ n =0 ; n ∈ N
x=(–2 ± √4+2π+8πn)/2
4+2π+8πn=4·(1+(π/2)+2πn)
x= –1± √1+(π/2)+2πn, n =0 ; n ∈ N ( или n ∈ N+)
N+ – 0 и натуральные.
О т в е т. –1± √1+(π/2)+2πn, n =0 ; n ∈ N