[tex](2sinx-1)(\sqrt{-cosx}+1)=0[/tex]
1.
одз:
-cosx≥0
cosx≤0
x=-3π/2+2πk
x=-π/2+2πk
x=[-3π/2+2πk;-π/2+2πk] . k=z
2.
1)
2sinx-1=0
sinx=1/2
x=π/6+2πk . k=z-не подх под одз
x=5π/6+2πk . k=z
2)
√-cosx+1=0
√-cosx=-1 корней нет
Ответ:x=5π/6+2πk . k=z
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Verified answer
1.
одз:
-cosx≥0
cosx≤0
x=-3π/2+2πk
x=-π/2+2πk
x=[-3π/2+2πk;-π/2+2πk] . k=z
2.
1)
2sinx-1=0
sinx=1/2
x=π/6+2πk . k=z-не подх под одз
x=5π/6+2πk . k=z
2)
√-cosx+1=0
√-cosx=-1 корней нет
Ответ:x=5π/6+2πk . k=z