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Evoc10
@Evoc10
August 2022
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a) 4cos^2 x = 3-4cosx
b)2sin^2 x + 3cosx -3 = 0
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sedinalana
Verified answer
А
4сos²x+4cosx-3=0
cosx=a
4a²+4a-3=0
D=16+48=64
a1=(-4-8)/8=-1,5⇒cosx=-1,5<-1 нет решения
a2=(-4+8)/8=1/2⇒cosx=1/2⇒x=+-π/3+2πn,n∈z
b
2-2cos²x+3cosx-3=0
cosx=a
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πn,n∈z
a2=(3+1)/4=1⇒cosx=1⇒x=2πk,k∈z
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Answers & Comments
Verified answer
А4сos²x+4cosx-3=0
cosx=a
4a²+4a-3=0
D=16+48=64
a1=(-4-8)/8=-1,5⇒cosx=-1,5<-1 нет решения
a2=(-4+8)/8=1/2⇒cosx=1/2⇒x=+-π/3+2πn,n∈z
b
2-2cos²x+3cosx-3=0
cosx=a
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πn,n∈z
a2=(3+1)/4=1⇒cosx=1⇒x=2πk,k∈z