$sin^2x+sin^2y+sin^2(x+y)=\frac{9}{4}$
Edited by trambau, 17-09-2017 - 20:29.
$sin^2x+sin^2y+sin^2(x+y)=\frac{9}{4}$
Edited by trambau, 17-09-2017 - 20:29.
$sin^2x+sin^2y+sin^2(x+y)=\frac{9}{4}$
$PT\Leftrightarrow \frac{1-cos2x}{2}+\frac{1-cos2y}{2}+sin^2(x+y)=\frac{9}{4}$
$\Leftrightarrow 1-cos(x+y)cos(x-y)+1-cos^2(x+y)=\frac{9}{4}$
$\Leftrightarrow cos^2(x+y)+cos(x+y)cos(x-y)+\frac{1}{4}=0$
ta có : $\Delta =cos^2(x-y)-1\leq 0$
$x,y$ là nghiệm của PT nên
$\Rightarrow \left\{\begin{matrix} cos^2(x-y)-1=0 & & \\ cos(x+y)=-\frac{cos(x-y)}{2} & & \end{matrix}\right. \Rightarrow \left\{\begin{matrix} cos(x-y)= \pm 1& & \\ cos(x+y)=-\frac{cos(x-y)}{2} & & \end{matrix}\right.$
từ đây giải các hệ ra nhé bạn
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