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onlyforalena
@onlyforalena
August 2022
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помогите (на фото)
√sin2x=√2*√cosx
промежуток |-3p/2;0|
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Answers & Comments
teddybrown05
Sin^2x=1-cos^2x->
y=cosx
y^2+2y-3=0
y1=-3-не удовлетворяет, так как [cosx]<=1
y2=1
x=2pi*n
На отрезке [-5pi;3pi]
x1=-4pi
x2=-2pi
x3=0
x4=2pi
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Answers & Comments
y=cosx
y^2+2y-3=0
y1=-3-не удовлетворяет, так как [cosx]<=1
y2=1
x=2pi*n
На отрезке [-5pi;3pi]
x1=-4pi
x2=-2pi
x3=0
x4=2pi