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vasilya1999
@vasilya1999
July 2022
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Решите тригонометрическое уравнение :
2x² sinx - 8sinx +4 =x²
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hlopushinairina
2x²sinx-8sinx+4=x²;
x²(2sinx-1)-4(2sinx-1)=0;
(x²-4)(2sinx-1)=0;
x²-4=0;
x=4;
x=-4;
2sinx-1=0;
sinx=1/2;
x=(-1)ⁿ·π/6+nπ;n∈Z
2 votes
Thanks 1
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Answers & Comments
x²(2sinx-1)-4(2sinx-1)=0;
(x²-4)(2sinx-1)=0;
x²-4=0;
x=4;
x=-4;
2sinx-1=0;
sinx=1/2;
x=(-1)ⁿ·π/6+nπ;n∈Z