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Alamart
@Alamart
August 2022
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Решите уравнение sinx= -√2/2 для cosx>0
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sedinalana
Verified answer
Cosx>0⇒x∈(-π/2+2πn;π/2+2πn,n∈z)
sinx=-√2/2
x=-π/4+2πn,n∈z
sinx=5π/4+2πn,n∈z не удов усл
7 votes
Thanks 6
petushkov20031
√2sinx(2+cosx)-(cosx+2)=0
(cosx+2)(√2sinx-1)=0
cosx=-2∉[-1;1]
sinx=1/√2⇒x=(-1)^n*π/4+πn
1 votes
Thanks 2
sultanovaadelin2
sinx=-√2/2
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Answers & Comments
Verified answer
Cosx>0⇒x∈(-π/2+2πn;π/2+2πn,n∈z)sinx=-√2/2
x=-π/4+2πn,n∈z
sinx=5π/4+2πn,n∈z не удов усл
(cosx+2)(√2sinx-1)=0
cosx=-2∉[-1;1]
sinx=1/√2⇒x=(-1)^n*π/4+πn