помогите, пожалуйста, очень срочно
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  • dubovik1999
    или же:
  • dubovik1999
    1-2sin^2(2x)+sin2x-1=0, sin2x=t,
    2t^2-t=0, t(2t-1)=0, t1=0, t2=1/2
    t1=0: sin2x=0, 2x=Пn, x=Пn/2.
    t2=1/2, sin2x=1/2, 2x=(-1)^k*П/6+Пk, x=(-1)^k*П/12+Пk/2

0 votes Thanks 1
dubovik1999 или же:
dubovik1999 1-2sin^2(2x)+sin2x-1=0, sin2x=t,
2t^2-t=0, t(2t-1)=0, t1=0, t2=1/2
t1=0: sin2x=0, 2x=Пn, x=Пn/2.
t2=1/2, sin2x=1/2, 2x=(-1)^k*П/6+Пk, x=(-1)^k*П/12+Пk/2
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