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921063
@921063
July 2022
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Найдите сумму корней уравнения cos^2(x)+sin(x)cos(x)=1, принадлежащих промежутку [-320градусов;50градусов)
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ХеленаВойцеховская
Вроде все перепроверила..............................
2 votes
Thanks 1
sedinalana
Cos²+sinxcosx-1=0
sinxcosx-sin²x=0
sinx(cosx-sinx)=0
sinx=0⇒x=πk,k∈z
cosx-sinx=0/cosx
1-tgx=0
tgx=1
x=π/4+πk,k∈z
-320≤180k<50
-320/180≤k<50/180
k=-1 x=-180
k=0 x=0
-320≤45+180k<50
-365≤180k<5
-365/180≤k≤5/180
k=-2 x=45-360=-315
k=-1 x=45-180=-135
k=0 x=45
3 votes
Thanks 3
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Answers & Comments
sinxcosx-sin²x=0
sinx(cosx-sinx)=0
sinx=0⇒x=πk,k∈z
cosx-sinx=0/cosx
1-tgx=0
tgx=1
x=π/4+πk,k∈z
-320≤180k<50
-320/180≤k<50/180
k=-1 x=-180
k=0 x=0
-320≤45+180k<50
-365≤180k<5
-365/180≤k≤5/180
k=-2 x=45-360=-315
k=-1 x=45-180=-135
k=0 x=45