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WinterWhite
@WinterWhite
July 2022
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Решите уравнение корень из 3 *sin2x+cos^2x=sin^2x
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zarembo73
Verified answer
√3sin2x+cos²x=sin²
x;
√3sin2x+(cos²x-sinx)=0;
√3sin2x+cos2x=0; |:2
√3/2sin2x+1/2cos2x=0;
√3/2=cosπ/6; 1/2=sinπ/6;
cosπ/6*sin2x+sinπ/6*cos2x=0; (формула сложения - сумма аргументов)
sin(2x+π/6)=0;
2x+π/6=πn, n∈Z;
2x=-π/6+πn, n∈Z;
x=-π/12+πn/2, n∈Z.
Ответ: -π/12+πn/2, n∈Z.
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Answers & Comments
Verified answer
√3sin2x+cos²x=sin²x;√3sin2x+(cos²x-sinx)=0;
√3sin2x+cos2x=0; |:2
√3/2sin2x+1/2cos2x=0;
√3/2=cosπ/6; 1/2=sinπ/6;
cosπ/6*sin2x+sinπ/6*cos2x=0; (формула сложения - сумма аргументов)
sin(2x+π/6)=0;
2x+π/6=πn, n∈Z;
2x=-π/6+πn, n∈Z;
x=-π/12+πn/2, n∈Z.
Ответ: -π/12+πn/2, n∈Z.