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marinamorenche
@marinamorenche
July 2022
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Решите уравнение : 1) sin 2x=3 cosx; 2)cosx+cos 2x=0; 3) 2 cosx+sinx=0
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sedinalana
Verified answer
1
2sinxcosx-3cosx=0
cosx(2sinx-3)=0
cosx=0⇒x=π/2+πn,n∈z
sinx=1,5>1 нет решения
2
cosx+2cos²x-1=0
cosx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒cosx=-1⇒x=π+2πn,n∈z
a2=(-1+3)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πk,k∈z
3
2cosx+sinx=0/cosx
2+tgx=0
tgx=-2
x=-arctg2+πn,n∈z
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Answers & Comments
Verified answer
12sinxcosx-3cosx=0
cosx(2sinx-3)=0
cosx=0⇒x=π/2+πn,n∈z
sinx=1,5>1 нет решения
2
cosx+2cos²x-1=0
cosx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒cosx=-1⇒x=π+2πn,n∈z
a2=(-1+3)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πk,k∈z
3
2cosx+sinx=0/cosx
2+tgx=0
tgx=-2
x=-arctg2+πn,n∈z