125 б и в пожалуйста))
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Ответил Artem112
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125б
![sin^2x-2cos^2x=sin2x
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sin^2x-2sin xcos x-2cos^2x=0
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mathrm{tg}^2x-2mathrm{tg}x-2=0
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D_1=(-1)^2-1cdot(-2)=3
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mathrm{tg}x= 1pm sqrt{3} \ x=mathrm{arctg}(1pm sqrt{3} )+ pi n. nin Z](https://tex.z-dn.net/?f=sin%5E2x-2cos%5E2x%3Dsin2x%0A%5C%0Asin%5E2x-2sin+xcos+x-2cos%5E2x%3D0%0A%5C%0Amathrm%7Btg%7D%5E2x-2mathrm%7Btg%7Dx-2%3D0%0A%5C%0AD_1%3D%28-1%29%5E2-1cdot%28-2%29%3D3%0A%5C%0Amathrm%7Btg%7Dx%3D+1pm+sqrt%7B3%7D+%5C+x%3Dmathrm%7Barctg%7D%281pm+sqrt%7B3%7D+%29%2B+pi+n.++nin+Z "sin^2x-2cos^2x=sin2x
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sin^2x-2sin xcos x-2cos^2x=0
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mathrm{tg}^2x-2mathrm{tg}x-2=0
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D_1=(-1)^2-1cdot(-2)=3
\
mathrm{tg}x= 1pm sqrt{3} \ x=mathrm{arctg}(1pm sqrt{3} )+ pi n. nin Z")
125в
![1+3sin^2x=2sin2x
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sin^2x+cos^2x+3sin^2x-2sin2x=0
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4sin^2x-4sin xcos x+cos^2x=0
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4mathrm{tg}^2x-4mathrm{tg}x+1=0
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(2mathrm{tg}x-1)^2=0
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2mathrm{tg}x-1=0
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mathrm{tg}x= frac{1}{2}
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x=mathrm{arctg}frac{1}{2} + pi n, nin Z](https://tex.z-dn.net/?f=1%2B3sin%5E2x%3D2sin2x%0A%5C%0Asin%5E2x%2Bcos%5E2x%2B3sin%5E2x-2sin2x%3D0%0A%5C%0A4sin%5E2x-4sin+xcos+x%2Bcos%5E2x%3D0%0A%5C%0A4mathrm%7Btg%7D%5E2x-4mathrm%7Btg%7Dx%2B1%3D0%0A%5C%0A%282mathrm%7Btg%7Dx-1%29%5E2%3D0%0A%5C%0A2mathrm%7Btg%7Dx-1%3D0%0A%5C%0Amathrm%7Btg%7Dx%3D+frac%7B1%7D%7B2%7D+%0A%5C%0Ax%3Dmathrm%7Barctg%7Dfrac%7B1%7D%7B2%7D+%2B+pi+n%2C++nin+Z "1+3sin^2x=2sin2x
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sin^2x+cos^2x+3sin^2x-2sin2x=0
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4sin^2x-4sin xcos x+cos^2x=0
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4mathrm{tg}^2x-4mathrm{tg}x+1=0
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(2mathrm{tg}x-1)^2=0
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2mathrm{tg}x-1=0
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mathrm{tg}x= frac{1}{2}
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x=mathrm{arctg}frac{1}{2} + pi n, nin Z")
125в
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