1.log (base 3) x - log (base x) 27 =2
Find x
2.4(sin x)^2+(sin x)(cos x)+3(cos x)^2=3
Find tan x
3.6(sin x)(cos x)+3sin x-2cos x-1=0
Find x
log+ trigo question F4
2012-08-21 5:30 am
回答 (2)
2012-08-21 6:30 am
✔ 最佳答案
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參考: My Maths World
2012-08-21 6:09 am
1) log3_x - logx_27 = 2
logx/log3 - 3log3/logx = 2
[(logx)^2 - 3(log3)^2]/(log3 logx) = 2
(logx)^2 -3(log3)^2 = 2log3 logx
(logx)^2 - log9 lox - 3(log3)^2 = 0
logx = ( log9 +/- sq.root( (log9)^2 + 12(log3)^2 )/2
= 3log3 or -log3
x = 27 or 1/3
2) 4(sin x)^2+(sin x)(cos x)+3(cos x)^2=3
sin^2(x) + (sinx)(cosx) + 3 - 3sin^2(x) - 3 = 0
2sin^2(x) -(sinx)(cosx) = 0
(sinx)(2sinx -cosx) = 0
2sinx - cosx = 0 or sinx = 0
2sinx = cosx
tanx = 1/2
3) 6(sinx)(cosx) + 3sinx - 2cosx -1 = 0
3sinx(2cosx +1) -( 2cosx +1)= 0
(3sinx-1) (2cosx +1)= 0
sinx = 1/3 or cosx = -1/2
x = arcsin(1/3) or x = 120+/-360n(度) or 240+/-360n(度) (由於x沒有範圍)
logx/log3 - 3log3/logx = 2
[(logx)^2 - 3(log3)^2]/(log3 logx) = 2
(logx)^2 -3(log3)^2 = 2log3 logx
(logx)^2 - log9 lox - 3(log3)^2 = 0
logx = ( log9 +/- sq.root( (log9)^2 + 12(log3)^2 )/2
= 3log3 or -log3
x = 27 or 1/3
2) 4(sin x)^2+(sin x)(cos x)+3(cos x)^2=3
sin^2(x) + (sinx)(cosx) + 3 - 3sin^2(x) - 3 = 0
2sin^2(x) -(sinx)(cosx) = 0
(sinx)(2sinx -cosx) = 0
2sinx - cosx = 0 or sinx = 0
2sinx = cosx
tanx = 1/2
3) 6(sinx)(cosx) + 3sinx - 2cosx -1 = 0
3sinx(2cosx +1) -( 2cosx +1)= 0
(3sinx-1) (2cosx +1)= 0
sinx = 1/3 or cosx = -1/2
x = arcsin(1/3) or x = 120+/-360n(度) or 240+/-360n(度) (由於x沒有範圍)
參考: 神風亨
收錄日期: 2021-04-13 18:55:42
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