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NataschaBro
@NataschaBro
July 2022
2
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Помогите решить
3sin2x-3cosx+2sinx-1=0
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Answers & Comments
dilara2504
3*2*sin(x)*cos(x) - 3*cos(x) + 2sin(x ) - 1 = 0
2*sin(x)*(3*cos(x) +1) - 1*(3*cos(x) + 1) = 0
(3*cos(x)+1) * (2*sin(x) -1) = 0
3*cos(x)+1= 0 ; 2*sin(x)-1 = 0
cos(x) = -1/3 ; sin(x) = 1/2
0 votes
Thanks 1
iqLL
Разложим по формуле синуса двойного угла и вынесем 3cosx
3cosx(2sinx-1) +(2sinx-1)=0
Можно еще вынести
(2sinx-1)(3cosx+1)=0
sinx=1/2
cosx=-1/3
x=(-1)^k pi/6 + pin
x=+-arccos(-1/3) +2pin
3 votes
Thanks 2
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Answers & Comments
2*sin(x)*(3*cos(x) +1) - 1*(3*cos(x) + 1) = 0
(3*cos(x)+1) * (2*sin(x) -1) = 0
3*cos(x)+1= 0 ; 2*sin(x)-1 = 0
cos(x) = -1/3 ; sin(x) = 1/2
3cosx(2sinx-1) +(2sinx-1)=0
Можно еще вынести
(2sinx-1)(3cosx+1)=0
sinx=1/2
cosx=-1/3
x=(-1)^k pi/6 + pin
x=+-arccos(-1/3) +2pin