Помогите решить,пожалуйста)))[tex] \sqrt{1,5 sin x} =cos x[/tex]. Created by Endddddd. algebra-ru

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July 2022 1 4 Report

Помогите решить,пожалуйста)))[tex] \sqrt{1,5 sin x} =cos x[/tex]

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Lesben √1,5sinx = cosx /²
1,5sin=cos²x
1,5sinx=1-sin²x
sin²ˇx+1,5sinx -1=0
sinx=z
z²+1,5z-1=0
D=2,25+4=6,25, √D=√6,25=2,5
z1=(-1,5+2,5)/2=1/2
z2=(-1,5-2,5)/2=-4/2=-2
a)sinx=1/2
x1=π/6+2k.π
x2=5π/6+2k.π , k∈Z
b)sinx=-2 , x∈∅

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