phuong trinh luong giac co ket hop ngiem

N

newstarinsky

2)ĐK $cosx\not=0$
PT trở thành
$cos^2x.(5sinx-2)=3(1-sinx).sin^2x\\
\Leftrightarrow (1-sinx)(1+sinx)(5sinx-2)=3sin^2x.(1-sinx)\\
\Leftrightarrow (1-sinx)(2sin^2x+3sinx-2)=0$

1) ĐK.....................
PT trở thành
$cosx-sin2x=\sqrt{3}+\sqrt{3}sinx-2\sqrt{3}sin^2x\\
\Leftrightarrow cosx-\sqrt{3}sinx=sin2x+\sqrt{3}(1-2sin^2x)=sin2x+\sqrt{3}cos2x\\
\Leftrightarrow sin(\dfrac{\pi}{6}-x)=sin(2x+\dfrac{\pi}{3})$
 
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