Предмет: Алгебра,
автор: zayaisraya01
в)-2cos^2x-5sinx-1=0
г)sin^2x+4sinxcosx-5cos^2=0
Ответы
Автор ответа:
0
в)
![-2cos^2x-5sinx-1=0
\-2(1-sin^2x)-5sinx-1=0
\sinx=y, y in [-1;1]
\-2+2y^2-5y-1=0
\2y^2-5y-3=0
\D=25+24=49=7^2
\y_1= frac{5+7}{4} =3 notin[-1;1]
\y_2= frac{5-7}{4} =- frac{1}{2} in [-1;1]
\sinx=- frac{1}{2}
\x_1= -frac{pi}{6} +2pi n, n in Z
\x_2=-frac{5pi}{6} +2pi n, n in Z](https://tex.z-dn.net/?f=-2cos%5E2x-5sinx-1%3D0%0A%5C-2%281-sin%5E2x%29-5sinx-1%3D0%0A%5Csinx%3Dy%2C+y+in+%5B-1%3B1%5D%0A%5C-2%2B2y%5E2-5y-1%3D0%0A%5C2y%5E2-5y-3%3D0%0A%5CD%3D25%2B24%3D49%3D7%5E2%0A%5Cy_1%3D+frac%7B5%2B7%7D%7B4%7D+%3D3+notin%5B-1%3B1%5D%0A%5Cy_2%3D+frac%7B5-7%7D%7B4%7D+%3D-+frac%7B1%7D%7B2%7D+in+%5B-1%3B1%5D%0A%5Csinx%3D-+frac%7B1%7D%7B2%7D%0A%5Cx_1%3D+-frac%7Bpi%7D%7B6%7D+%2B2pi+n%2C++n+in+Z%0A%5Cx_2%3D-frac%7B5pi%7D%7B6%7D+%2B2pi+n%2C++n+in+Z "-2cos^2x-5sinx-1=0
\-2(1-sin^2x)-5sinx-1=0
\sinx=y, y in [-1;1]
\-2+2y^2-5y-1=0
\2y^2-5y-3=0
\D=25+24=49=7^2
\y_1= frac{5+7}{4} =3 notin[-1;1]
\y_2= frac{5-7}{4} =- frac{1}{2} in [-1;1]
\sinx=- frac{1}{2}
\x_1= -frac{pi}{6} +2pi n, n in Z
\x_2=-frac{5pi}{6} +2pi n, n in Z")
г)
г)
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