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ninareyzvih
@ninareyzvih
August 2022
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Решите уравнение
2cos^2 x=3sinx+2
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rairura
2(1-sinx^2)-3sinx-2=0
2-2sin^2x-3sinx-2=0
-2sin^2x-3sinx=0
sinx(2sinx+3)=0
sinx=0; sinx=-3/2
x=Пn; x=(-1)^n+1*arcsin 3/2 +Пn
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Answers & Comments
2-2sin^2x-3sinx-2=0
-2sin^2x-3sinx=0
sinx(2sinx+3)=0
sinx=0; sinx=-3/2
x=Пn; x=(-1)^n+1*arcsin 3/2 +Пn