### + / -檢視主題

#ed_op#P#ed_cl#應該是沒設 x 的範圍吧! 如果有設 0 ≦ x &lt; 2#ed_op#FONT face="Comic Sans MS"#ed_cl#π#ed_op#/FONT#ed_cl#, 就是兩個解了! 如圖#ed_op#/P#ed_cl##ed_op#P#ed_cl##ed_op#IMG alt="image file name: 2k37b240c483.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k37b240c483.png" border=0#ed_cl##ed_op#/P#ed_cl##ed_op#P#ed_cl#3cos(x) + 4sin(x) = 2#ed_op#/P#ed_cl##ed_op#P#ed_cl#設 s = sin(x)#ed_op#/P#ed_cl##ed_op#P#ed_cl#3√(1-s#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#) + 4s = 2#ed_op#BR#ed_cl#4s - 2 =&nbsp; - 3√(1-s#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#)#ed_op#BR#ed_cl#(4s - 2)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# =&nbsp; (- 3√(1-s#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#))#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/P#ed_cl##ed_op#P#ed_cl#25s#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# - 16s - 5 = 0#ed_op#BR#ed_cl#s = (8-3√21)/25, (8+3√21)/25 = sin(a), sin(b) ..#ed_op#/P#ed_cl##ed_op#P#ed_cl#cos(a) = √(1-sin#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#(a)) #ed_op#BR#ed_cl#= √(1-((8-3√21)/25)^2) #ed_op#BR#ed_cl#= (4√21 +6)/25#ed_op#/P#ed_cl##ed_op#P#ed_cl#cos(b) = √(1-sin#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#(b)) #ed_op#BR#ed_cl#= √(1-((8+3√21)/25)^2) #ed_op#BR#ed_cl#= (4√21 -6)/25#ed_op#/P#ed_cl##ed_op#P#ed_cl#sin(a+b) = sin(a)cos(b) + sin(b)cos(a)#ed_op#BR#ed_cl#= ((8-3√21)/25)((4√21 -6)/25) + ((8+3√21)/25)((4√21 +6)/25)#ed_op#BR#ed_cl#= 4√21/25#ed_op#/P#ed_cl##ed_op#P#ed_cl#cos(a+b) = cos(a)cos(b) - sin(a)sin(b)#ed_op#BR#ed_cl#= ((4√21 +6)/25)((4√21 -6)/25) + ((8-3√21)/25)((8+3√21)/25)#ed_op#BR#ed_cl#= 7/25#ed_op#/P#ed_cl##ed_op#P#ed_cl#tan(t/2) = (1 - cos(t)) /sin(t)#ed_op#/P#ed_cl##ed_op#P#ed_cl#tan((a+b)/2) = (1 - cos(a+b)) / sin(a+b)#ed_op#BR#ed_cl#= (1 - (7/25)) / (4√21/25)#ed_op#BR#ed_cl#= 3·√21 / 14#ed_op#/P#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

### [問題]題目是否有誤

#ed_op#DIV#ed_cl##ed_op#FONT face=Verdana#ed_cl#a,b為4sinx+3cosx=2之兩相異根,求tan[(a+b)/2]=?#ed_op#BR#ed_cl##ed_op#BR#ed_cl#各位會不會覺得題目怪怪的?#ed_op#BR#ed_cl##ed_op#BR#ed_cl#這題x有無限多解#ed_op#BR#ed_cl##ed_op#BR#ed_cl#a,b是無限多解中之相異兩個?? #ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl#