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à¤Ã×èͧÁ×ͧ͢ËÑÇ¢éÍ | ¤é¹ËÒã¹ËÑÇ¢é͹Õé |
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#1
18 ÊÔ§ËÒ¤Á 2008, 20:14
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µÃÕ⡳«Ñ¡¹Ô´¤ÃѺ
¶éÒ $\cos{3x}=\sin{x}$ ¨§ËÒ¤èҢͧx
__________________
I think you're better than you think you are. 
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#2
18 ÊÔ§ËÒ¤Á 2008, 20:28
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·ÓÍÂèÒ§¹Õéä´éËÃ×Íà»ÅèÒ
$$cos3x=sinx$$ $$cos3x=cos(\frac{\pi }{2} -x)$$ $$3x=\frac{\pi}{2} -x$$ $$x= \frac{\pi }{8} +2n\pi $$ 18 ÊÔ§ËÒ¤Á 2008 22:25 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ square1zoa à˵ؼÅ: äÁèá¡é´Õ¡ÇèÒ ¼Ô´à»ç¹Á¹ØÉÂì
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#3
18 ÊÔ§ËÒ¤Á 2008, 20:52
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¼Ô´¤ÃѺ ¤ÓµÍº¢Í§¢é͹Õé¤×Í $x=\frac{(4k+1)\pi}{8},\frac{(4k+3)\pi}{4}$ â´Â·Õè $k\in\mathbb{Z}$ ¤ÃѺ
¤Ø³ square1zoa ¼Ô´µÃ§¨Ò¡ºÃ÷Ѵ·Õè 2 ä» 3 ¤ÃѺ ÁѹäÁèÊÒÁÒà imply à»ç¹ºÃ÷Ѵ·Õè 3 㹷ѹ·Õ¤ÃѺ
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#4
18 ÊÔ§ËÒ¤Á 2008, 21:00
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§Ñé¹ãËé¤Ø³ owlpenguin ªèÇÂáÊ´§ÇÔ¸Õ·ÓãËé´Ù˹èͤÃѺ
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I think you're better than you think you are.
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#5
18 ÊÔ§ËÒ¤Á 2008, 21:21
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àÍèÍ ¼Ô´¨ÃÔ§´éÇ ¢ÍàÇÅÒ¹Ñ觤Դ¹Ð¤ÃѺ
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#6
18 ÊÔ§ËÒ¤Á 2008, 22:08
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·ÓẺ¹Õé ¶Ù¡ÍêлèÒǤѺ
 $$\sin(\frac{\pi }{2} -3x)-\sin(x)=0$$ $$2\cos(\frac{\pi }{4} -x)\sin(\frac{\pi }{4} -2x)=0$$ ¾Ô¨ÒóÒá¤è $$\cos(x-\frac{\pi }{4})\sin(2x-\frac{\pi }{4})=0$$ ¨Ðä´é $x-\frac{\pi }{4}=k\pi+\frac{2\pi }{4}$ áÅÐ $2x-\frac{\pi }{4}=k\pi$ $\therefore$ $ x=\frac{(4k+1)\pi}{8},\frac{(4k+3)\pi}{4}$
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I'm kak.
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#7
18 ÊÔ§ËÒ¤Á 2008, 22:25
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WOW! Å×ÁÊÙµÃä´éä§àÃÒà¹ÕèÂ
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