Question

一道简单的高一数学题求解

已知sin^2 2x+sin2xcosx-cos2x=1,x为锐角，求sinx,tanx
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急求!!!!!

Answer 1

sin^2 2x+sin2xcosx-cos2x=1 =>
sin^2 2x+sin2xcosx-(cos2x-1)=0 =>
sin^2 2x+sin2xcosx-2*cos^2 x=0 =>
(sin2x+2cosx)(sin2x-cosx)=0 =>
sin2x=2sinxcosx=-2cosx 或sin2x=2sinxcosx=cosx
由于x为锐角,所以2sinx=1,即x=Pi/6
因此sinx=1/2 tanx=(根号3)/3

Answer 2

展开
4sin^2 xcos^2 x+2sinxcos^2 x-cos^2 x+sin^2 x=cos^2 x+sin^2 x
化简得
2sin^2 x+sinx-1=0
所以
sinx=1/2
因为x是锐角，所以x=30°
tanx=sqr3/3