Ответ:
x₁=-π/2+2πk, k∈Z
x₂=±π/6+2πk, k∈Z
Объяснение:
Воспользуемся формулой приведения:
sin(π/3-x)=cos(π/2-(π/3-x))=cos(π/6+x)
2cos²(x+π/6)-3cos(x+π/6)+1=0
a=cos(x+π/6), -1≤a≤1
2a²-3a+1=0
D=3²-4*2=1
a₁=(3-1)/4=1/2
a₂=(3+1)/4=1 ⇒
cos(x+π/6)=1/2
x+π/6=±π/3+2πk, k∈Z
x₂=π/6+2πk, k∈Z
cos(x+π/6)=1
x+π/6=2πk, k∈Z
x₃=-π/6+2πk, k∈Z
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Verified answer
Ответ:
x₁=-π/2+2πk, k∈Z
x₂=±π/6+2πk, k∈Z
Объяснение:
Воспользуемся формулой приведения:
sin(π/3-x)=cos(π/2-(π/3-x))=cos(π/6+x)
2cos²(x+π/6)-3cos(x+π/6)+1=0
a=cos(x+π/6), -1≤a≤1
2a²-3a+1=0
D=3²-4*2=1
a₁=(3-1)/4=1/2
a₂=(3+1)/4=1 ⇒
cos(x+π/6)=1/2
x+π/6=±π/3+2πk, k∈Z
x₁=-π/2+2πk, k∈Z
x₂=π/6+2πk, k∈Z
cos(x+π/6)=1
x+π/6=2πk, k∈Z
x₃=-π/6+2πk, k∈Z