Question

9£®ÒÑÖªÏòÁ¿$\overrightarrow{m}$=£¨$\sqrt{3}$sin¦Øx-cos¦Øx£¬1£©£¬$\overrightarrow{n}$=£¨cos¦Øx£¬$\frac{1}{2}$£©£¬É躯Êýf£¨x£©=$\overrightarrow{m}•\overrightarrow{n}$£¬
Èôº¯Êýf£¨x£©µÄͼÏó¹ØÓÚÖ±Ïßx=$\frac{¦Ð}{3}$¶Ô³ÆÇҦءÊ[0£¬2]
£¨¢ñ£© Çóº¯Êýf£¨x£©µÄµ¥µ÷µÝ¼õÇø¼ä£»
£¨¢ò£© ÔÚ¡÷ABCÖУ¬½ÇA£¬B£¬CµÄ¶Ô±ß·Ö±ða£¬b£¬c£¬Èôa=$\sqrt{3}$£¬f£¨A£©=1£¬Çób+cµÄ×î´óÖµ£®

Answer 1

·ÖÎö £¨¢ñ£©»¯¼òf£¨x£©£¬ÀûÓÃÖÜÆÚ¹«Ê½Çó³ö¦ØµÃ³öf£¨x£©µÄ½âÎöʽ£¬ÀûÓÃÕýÏÒº¯ÊýµÄµ¥µ÷ÐÔÁгö²»µÈʽ½â³öµ¥µ÷ÔöÇø¼ä£»
£¨¢ò£©Í¨¹ýf£¨A£©=1£¬Çó³öAµÄÖµ£¬ÀûÓÃÓàÏÒ¶¨ÀíµÃµ½¹ØÓÚb+cµÄ±í´ïʽ£¬È»ºóÇóÆä×î´óÖµ£®

½â´ð ½â£º£¨¢ñ£©f£¨x£©=£¨$\sqrt{3}$sin¦Øx-cos¦Øx£©cos¦Øx+$\frac{1}{2}$=$\sqrt{3}$sin¦Øx•cos¦Øx-cos²¦Øx+$\frac{1}{2}$
=$\frac{\sqrt{3}}{2}sin2¦Øx$-$\frac{1}{2}cos2¦Øx$=sin£¨2¦Øx-$\frac{¦Ð}{6}$£©
º¯Êýf£¨x£©µÄͼÏó¹ØÓÚÖ±Ïßx=$\frac{¦Ð}{3}$¶Ô³Æ£¬Ôò$\frac{2¦Ø¦Ð}{3}-\frac{¦Ð}{6}=k¦Ð+\frac{¦Ð}{2}£¬k¡ÊZ$
Ôò$¦Ø=\frac{3}{2}k+1$£¬k¡ÊZÇҦءÊ[0£¬2]£¬Ôò¦Ø=1¡­£¨4·Ö£©
¡àf£¨x£©=sin£¨2x-$\frac{¦Ð}{6}$£©£¬Áî2k¦Ð+$\frac{¦Ð}{2}$¡Ü2x-$\frac{¦Ð}{6}$$¡Ü\frac{3¦Ð}{2}$+2k¦Ð£¬½âµÃk¦Ð+$\frac{¦Ð}{3}$$¡Üx¡Ük¦Ð+\frac{5¦Ð}{6}$£¬k¡ÊZ
¡àº¯Êýf£¨x£©µÄµ¥µ÷µÝ¼õÇø¼äΪ[k¦Ð+$\frac{¦Ð}{3}$£¬k¦Ð+$\frac{5¦Ð}{6}$]£¬k¡ÊZ£®
£¨¢ò£©f£¨A£©=sin£¨2A-$\frac{¦Ð}{6}$£©=1£¬ÇÒAÊÇ¡÷ABCÄڽǣ¬
¡à0£¼A£¼¦Ð£¬Ôò-$\frac{¦Ð}{6}$£¼2A-$\frac{¦Ð}{6}$$£¼\frac{11¦Ð}{6}$£¬ËùÒÔ2A-$\frac{¦Ð}{6}$=$\frac{¦Ð}{2}$£¬ÔòA=$\frac{¦Ð}{3}$£¬
¡ßa=$\sqrt{3}$£¬ÓÉÓàÏÒ¶¨Àí${a}^{2}={b}^{2}+{c}^{2}-2bc•cos\frac{¦Ð}{3}$=b²+c²-bc=£¨b+c£©²-3bc
Ôò£¨b+c£©²-3bc=3£¬¶øbc¡Ü£¨$\frac{b+c}{2}$£©²£¬ËùÒÔ3=£¨b+c£©²-3bc¡Ý£¨b+c£©²-3¡Á£¨$\frac{b+c}{2}$£©²=$\frac{£¨b+c£©^{2}}{4}$
⇒b+c$¡Ü2\sqrt{3}$£¬µ±ÇÒ½öµ±b=c=$\sqrt{3}$ʱ£¬
ËùÒÔb+cµÄ×î´óֵΪ2$\sqrt{3}$£®

µãÆÀ ±¾Ì⿼²éÈý½Çº¯ÊýµÄ»¯¼òÇóÖµ£¬½âÈý½ÇÐεÄ֪ʶ£¬¶þ±¶½Ç¹«Ê½¡¢Á½½ÇºÍµÄÕýÏÒº¯Êý¡¢ÓàÏÒ¶¨ÀíµÄÓ¦Ó㬿¼²é¼ÆËãÄÜÁ¦£¬×¢ÒâAµÄ´óСÇó½â£¬ÊÇÒ×´íµã£®