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1219али
@1219али
July 2022
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сколько корней уравнение sin x +1/2/cos(x+пи/3)=0 попадает в интервал [0;2пи]
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sedinalana
Verified answer
(sinx+1/2)/cos(x+π/3)=0
cos(x+π/3)≠0
x+π/3≠π/2+πn
x≠π/6+πn,n∈z
sinx+1/2=0
sinx=-1/2
x=-π/6+2πn,n∈z
x=7π/6+2πn,n∈z не удов усл
0≤-π/6+2πn≤2π
0≤-1+12n≤12
1≤12n≤13
1/12≤n≤13/12
n=1⇒x=-π/6+2π=11π/6∈[0;2π]
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Answers & Comments
Verified answer
(sinx+1/2)/cos(x+π/3)=0cos(x+π/3)≠0
x+π/3≠π/2+πn
x≠π/6+πn,n∈z
sinx+1/2=0
sinx=-1/2
x=-π/6+2πn,n∈z
x=7π/6+2πn,n∈z не удов усл
0≤-π/6+2πn≤2π
0≤-1+12n≤12
1≤12n≤13
1/12≤n≤13/12
n=1⇒x=-π/6+2π=11π/6∈[0;2π]