$\left\{\begin{matrix} sinx+siny=1 & (1) \\ cosx+cosy=\sqrt{3} & (2) \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} sin^{2}x+2sinxsiny+sin^{2}y=1 & \\ cos^{2}x+2cosxcosy+cos^{2}y=3 & \end{matrix}\right.$
$\Rightarrow sin^{2}x+2sinxsiny+sin^{2}y+ cos^{2}x+2cosxcosy+cos^{2}y=1+3$
$\Leftrightarrow (sin^{2}x+ cos^{2}x)+(sin^{2}y +cos^{2}y)+2(cosxcosy+sinxsiny)=4$
$\Leftrightarrow 1+1+2cos(x-y)=4 \Leftrightarrow 2cos(x-y)=2 \Leftrightarrow cos(x-y)=1 \Leftrightarrow x-y=k2\pi \Leftrightarrow x=y+k2\pi$ $(k \in \mathbb{Z})$
Khi đó $:$ $(1) \Leftrightarrow sin(y+k2\pi)+siny=1 \Leftrightarrow siny.cos(k2\pi)+cosy.sin(k2\pi)+siny=1 \Leftrightarrow siny.1+cosy.0+siny=1$
$\Leftrightarrow 2siny=1 \Leftrightarrow siny=\frac{1}{2}$ $(*)$
$(2) \Leftrightarrow cos(y+k2\pi)+cosy=\sqrt{3} \Leftrightarrow cosy.cos(k2\pi)-siny.sin(k2\pi)+cosy=\sqrt{3} \Leftrightarrow cosy.1-siny.0+cosy=\sqrt{3}$
$\Leftrightarrow 2cosy=\sqrt{3} \Leftrightarrow cosy=\frac{\sqrt{3}}{2}=cos\frac{\pi}{6} \Leftrightarrow y=\pm \frac{\pi}{6}+l2\pi$ $(l \in \mathbb{Z})$
Do $(*)$ nên $y= \frac{\pi}{6}+l2\pi \Rightarrow x= y+k2\pi= \frac{\pi}{6}+l2\pi+ k2\pi= \frac{\pi}{6}+(l+k)2\pi$
Vậy hệ phương trình có nghiệm $(x;y)=(\frac{\pi}{6}+(l+k)2\pi; \frac{\pi}{6}+l2\pi)$ $(k,l \in \mathbb{Z})$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
