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August 2022
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Легко плз. Найдите корни уравнения 6sin²x-5cosx-5=0 на промежутке [3Π; 5П]
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tseluyko58
Verified answer
6*(1-cos^2 x)-5cosx-5=0
-6cos^2 x-5cosx+1=0
6cos^2 x+5cos x-1=0
cos x=t, |t|<=1
6t^2+5t-1=0
t=1/6, t=-1
cos x=1/6, x=+-arccos1/6+2pin
cos x=-1, x=pi+2pin. x=3pi,5pi,+-arccos1/6+4pi
3 votes
Thanks 1
fb17
тут же нужно на промежутке
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Answers & Comments
Verified answer
6*(1-cos^2 x)-5cosx-5=0-6cos^2 x-5cosx+1=0
6cos^2 x+5cos x-1=0
cos x=t, |t|<=1
6t^2+5t-1=0
t=1/6, t=-1
cos x=1/6, x=+-arccos1/6+2pin
cos x=-1, x=pi+2pin. x=3pi,5pi,+-arccos1/6+4pi