Задача:

sin2x+1–2sin2x=0 1–2sin2×2x+sin2x=0 –sin2x+2sin2×2x–1=0 sin2x=t 2t2–t–1=0 D=(–1)2–4×2(–1)=9 x1,2=1+3/4 x1=1 x2=–1/2 sinx=1 x=n/2+2Пп sinx=–1/2 x=n/6–2Пm x2=5n/6–2nk