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namotrasnik
@namotrasnik
July 2022
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Помогите, пожалуйста, решить пример [tex]\frac{(6sin^2x-6sinx+cos2x+1)}{(12x^2-8 \pi x+ \pi ^2)}=0[/tex]
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sedinalana
Verified answer
12x²-8πx+π²≠0
D=64π²-48π²=16π²
x1≠(8π-4π)/24=π/6 Ux2≠(8π+4π)/24=π/2
6sin²x-6sinx+1-2sin²x+1=0
sinx=a
4a²-6a+2=0
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn U x≠π/6⇒x=5π/6+2πn,n∈z
a2=(3+1)/4=1⇒sinx=1⇒x=π/2+2πn U x≠π/2⇒x=5π/2+2πn,n∈z
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Answers & Comments
Verified answer
12x²-8πx+π²≠0D=64π²-48π²=16π²
x1≠(8π-4π)/24=π/6 Ux2≠(8π+4π)/24=π/2
6sin²x-6sinx+1-2sin²x+1=0
sinx=a
4a²-6a+2=0
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn U x≠π/6⇒x=5π/6+2πn,n∈z
a2=(3+1)/4=1⇒sinx=1⇒x=π/2+2πn U x≠π/2⇒x=5π/2+2πn,n∈z