Question

11£®ÒÑÖªÏòÁ¿$\overrightarrow m=£¨{2sinx£¬1}£©£¬\overrightarrow n=£¨{sinx+\sqrt{3}cosx£¬-3}£©£¬x¡ÊR$£¬º¯Êýf£¨x£©=$\overrightarrow{m}$•$\overrightarrow{n}$+2£®
£¨¢ñ£©Çóº¯Êýf£¨x£©µÄ×îСÕýÖÜÆÚ£»
£¨¢ò£©ÉèÈñ½Ç¡÷ABCÄÚ½ÇA£¬B£¬CËù¶ÔµÄ±ß·Ö±ðΪa£¬b£¬c£¬Èôf£¨A£©=2£¬$a=\sqrt{7}£¬b=3$£¬Çó½ÇAºÍ±ßcµÄÖµ£®

Answer 1

·ÖÎö £¨I£©ÀûÓÃƽÃæÏòÁ¿ÊýÁ¿»ýµÄÔËË㣬Èý½Çº¯ÊýºãµÈ±ä»»µÄÓ¦Óû¯¼òº¯Êý½âÎöʽ¿ÉµÃf£¨x£©=$2sin£¨2x-\frac{¦Ð}{6}£©$£¬ÀûÓÃÈý½Çº¯ÊýµÄÖÜÆÚ¹«Ê½¼´¿ÉµÃ½â£®
£¨II£©ÓÉ£¨I£©Öª¿ÉµÃ$sin£¨2A-\frac{¦Ð}{6}£©=1$£¬½áºÏAµÄ·¶Î§¿ÉÇó$A=\frac{¦Ð}{3}$£¬½â·¨Ò»£ºÓÉÓàÏÒ¶¨Àí½âµÃcµÄÖµ£¬½â·¨¶þ£ºÓÉÕýÏÒ¶¨Àí½âµÃsinB£¬ÓÉBÊÇÈñ½Ç£¬¿ÉÇócosB£¬ÀûÓÃÈý½ÇÐÎÄڽǺͶ¨Àí£¬Á½½ÇºÍµÄÕýÏÒº¯Êý¹«Ê½¿ÉÇósinC£¬¸ù¾ÝÕýÏÒ¶¨Àí¼´¿É½âµÃcµÄÖµ£®

½â´ð ½â£º£¨I£©f£¨x£©=$\overrightarrow{m}$•$\overrightarrow{n}$+2=$2sinx£¨sinx+\sqrt{3}cosx£©-3+2$¡­£¨2·Ö£©
=$2{sin^2}x+2\sqrt{3}sinxcosx-1$
=$\sqrt{3}sin2x-cos2x$¡­£¨4·Ö£©
=$2sin£¨2x-\frac{¦Ð}{6}£©$¡­£¨5·Ö£©
¡àf£¨x£©µÄ×îСÕýÖÜÆÚ$T=\frac{2¦Ð}{2}=¦Ð$¡­£¨6·Ö£©
£¨II£©ÓÉ£¨I£©Öª$f£¨A£©=2sin£¨2A-\frac{¦Ð}{6}£©=2$£¬½âµÃ$sin£¨2A-\frac{¦Ð}{6}£©=1$£®¡­£¨7·Ö£©
¡ß$A¡Ê£¨{0£¬\frac{¦Ð}{2}}£©£¬\;¡à2A-\frac{¦Ð}{6}¡Ê£¨{-\frac{¦Ð}{6}£¬\frac{5¦Ð}{6}}£©$£¬
¡à$2A-\frac{¦Ð}{6}=\frac{¦Ð}{2}$£¬¡à$A=\frac{¦Ð}{3}$£®¡­£¨9·Ö£©
½â·¨Ò»£ºÓÉÓàÏÒ¶¨ÀíµÃ${a^2}={3^2}+{c^2}-2¡Á3c¡Ácos\frac{¦Ð}{3}$=c²-3c+9=7£®
½âµÃc=1»òc=2£®¡­£¨10·Ö£©
Èôc=1£¬Ôò$cosB=\frac{{{1^2}+{{£¨{\sqrt{7}}£©}^2}-{3^2}}}{{2¡Á1¡Á\sqrt{7}}}$£¼0£¬
¡àBΪ¶Û½Ç£¬ÕâÓë¡÷ABCΪÈñ½ÇÈý½ÇÐβ»·û£¬c¡Ù1£®¡­£¨11·Ö£©
¡àc=2£®¡­£¨12·Ö£©
½â·¨¶þ£ºÓÉÕýÏÒ¶¨ÀíµÃ$\frac{3}{sinB}=\frac{{\sqrt{7}}}{{sin\frac{¦Ð}{3}}}$£¬½âµÃ$sinB=\frac{{3\sqrt{21}}}{14}$¡­£¨10·Ö£©
¡ßBÊÇÈñ½Ç£¬¡à$cosB=\sqrt{1-{{sin}^2}B}=\frac{{\sqrt{7}}}{14}$£¬
¡ßC=¦Ð-£¨A+B£©£¬
¡à$sinC=sin£¨A+B£©=sin£¨\frac{¦Ð}{3}+B£©=\frac{{\sqrt{21}}}{7}$£¬¡­£¨11·Ö£©
¡à$\frac{c}{{\frac{{\sqrt{21}}}{7}}}=\frac{{\sqrt{7}}}{{\frac{{\sqrt{3}}}{2}}}$£¬½âµÃc=2£®¡­£¨12·Ö£©

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËƽÃæÏòÁ¿ÊýÁ¿»ýµÄÔËË㣬Èý½Çº¯ÊýºãµÈ±ä»»µÄÓ¦Óã¬Èý½Çº¯ÊýµÄÖÜÆÚ¹«Ê½£¬ÓàÏÒ¶¨Àí£¬ÕýÏÒ¶¨Àí£¬Èý½ÇÐÎÄڽǺͶ¨Àí£¬Á½½ÇºÍµÄÕýÏÒº¯Êý¹«Ê½ÔÚ½âÈý½ÇÐÎÖеÄÓ¦Ó㬿¼²éÁËת»¯Ë¼ÏëºÍÊýÐνáºÏ˼ÏëµÄÓ¦Ó㬿¼²éÁ˼ÆËãÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮