а)корень из 2 на cos^2=sin(x-pi/2) б) отметить точки на промежутке от [-3п/2;-п]. Created by filiaa. algebra-ru

2;-п]...

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August 2022 1 3 Report

а)корень из 2 на cos^2=sin(x-pi/2)
б) отметить точки на промежутке от [-3п/2;-п]

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sedinalana

Verified answer

√2cos²x+cosx=0
cosx(√2cosx+1)=0
cosx=0
x=π/2+πn,n∈z
-3π/2≤π/2+πn≤-π
-3≤1+2n≤-2
-4≤2n≤-3
-2≤n≤-1,5
n=-2⇒x=π/2-2π=-3π/2
√2cosx+1=0
cosx=-1/√2
x=-3π/4+2πk,k∈z U x=3π/4+2πt,t∈z
-3π/2≤-3π/4+2πk≤-π
-6≤-3+8k≤-4
-3≤8k≤-1
-3/8≤k≤-1/8
нет решения
-3π/2≤3π/4+2πt≤-π
-6≤3+8t≤-4
-9≤8t≤-7
-9/8≤t≤-7/8
t=-1⇒x=3π/4-2π=-5π/4

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Answers & Comments

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2podrobno pliz

pomogite najti primery s grammaticheskoj oshibkoj

x 2 x gt0 pomogite s predelami

gelevaya ruchka stoit 60 rublej kakoe naibolshee chislo takih ruchek mozhno budet ku

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