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madamsshsmsnv
@madamsshsmsnv
August 2022
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3 sin в квадрате x - 4sinxcosx+cos в квадрате x=0
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EgorTat
3(sinx)^2-4sinxcosx+(cosx)^2=0; 2(sinx)^2-4sinxcosx+1=0;
2sinx(sinx-2cosx)+1=0; 2sinx(sinx -2(1-(sinx)^2)+1=0;
2(sinx)^2-4sinx+4(sinx)^3+1=0; 4y^3+2y^2-4y+1=0
x=(tg1/3)^(-1)+pn, n Î
Z; x=1/4(
p+4pn), n Î
Z
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Answers & Comments
2sinx(sinx-2cosx)+1=0; 2sinx(sinx -2(1-(sinx)^2)+1=0;
2(sinx)^2-4sinx+4(sinx)^3+1=0; 4y^3+2y^2-4y+1=0
x=(tg1/3)^(-1)+pn, n ÎZ; x=1/4(p+4pn), n ÎZ