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claic
@claic
July 2022
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Решение уравнения: 3cos^2x-sinx-1=0
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mikael2
Verified answer
3(1-sin²x)-sinx-1= -3sin²x-sinx+2=0
3sin²x+sinx-2=0 D=1+24=25 √D=5
sinx=(-1+5)/6=2/3 x=(-1)^n *arcsin 2/3+πn n∈Z
sinx=(-1-5)/6= -1 x=3/2π+2πn n∈Z
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Answers & Comments
Verified answer
3(1-sin²x)-sinx-1= -3sin²x-sinx+2=03sin²x+sinx-2=0 D=1+24=25 √D=5
sinx=(-1+5)/6=2/3 x=(-1)^n *arcsin 2/3+πn n∈Z
sinx=(-1-5)/6= -1 x=3/2π+2πn n∈Z