標題:

F.4 MATHS

發問:

3sin2θ=2 , 0°<θ<45° ----------------------- 2tan(2x-17°)=1 , 8.5°<53.5° ----------------------- 3logx+log1/4=logx2-log3 ----------------------- logx2=(logx)2 ----------------------- log(3x-4)+log2=2logx 希望有人可以解答到我哩d問題>..<... THZ...

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最佳解答:

( 1 ) 3 sin 2θ = 2 sin 2θ = 2 / 3 2θ = 41.8103*, 138.1897* θ = 20.9* ( cor. to 3 s.f. ) or θ = 69.1* ( rejected as 0 < θ < 45 ) Hence the solution is θ = 20.9* ( 2 ) 2 tan(2x-17°) = 1 tan(2x-17°) = 1 / 2 2x - 17°= 26.5651* or 206.5651* 2x = 43.5651* or 223.5651* x = 21.8* ( cor. to 3 s.f. ) or 112* ( rejected as 8.5 < θ < 53.5 ) Hence the solution is x = 21.8*. ( 3 ) 3logx + log1/4 = logx2 - log3 log ( x^3 ) + log ( 1 / 4 ) = log ( x^2 / 3 ) log ( x^3 / 4 ) = log ( x^2 / 3 ) x^3 / 4 = x^2 / 3 3x^3 - 4x^2 = 0 x^2 ( 3x - 4 ) = 0 x = 0 ( rejected ) or x = 4 / 3 ( 4 ) logx2=(logx)2 2 log x = ( log x )^2 ( log x )^2 - 2 log x = 0 log x ( log x - 2 ) = 0 log x = 0 or log x = 2 x = 1 or 100 ( 5 ) log(3x-4) + log2 = 2logx log 2 ( 3x - 4 ) = log ( x^2 ) 2 ( 3x - 4 ) = x^2 6x - 8 = x^2 x^2 - 6x + 8 = 0 ( x - 4 )( x - 2 ) = 0 x = 4 or 2

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