求方程cos^2x-sin^2=1/2在区间[-2π,2π]上所有的解的和

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求方程cos^2x-sin^2=1/2在区间[-2π,2π]上所有的解的和
求方程cos^2x-sin^2=1/2在区间[-2π,2π]上所有的解的和

求方程cos^2x-sin^2=1/2在区间[-2π,2π]上所有的解的和
cos^2x-sin^2x=cos(2x)=1/2
解有x=2kπ±π/3,k∈Z
故当k=-2时x1=-11π/6
当k=-1时x2=-7π/6,x3=-5π/6
当k=0时x4=-π/6,x5=π/6
当k=1时x6=5π/6,x7=7π/6
当k=2时x8=11π/6
所以x在[-2π,2π]上所有解的和:x1+x2+x3+x4+x5+x6+x7+x8=0

cos^2X-sin^2X=1/2=cos2X
2X=2kπ±π/3
X=kπ±π/6
k=-2,X1=-2π+π/6=-11π/6
k=-1,X2=-π-π/6=-7π/6
X3=-π+π/6=-5π/6
k=0 ,X4=π/6
X5=-π/6
k=1,X6=π-π/6=5π/6
X7=π+π/6=7π/6
k=2,X8=2π-π=11π/6

题目应该是:cos^2x-sin^2x =1/2.吧,
解题如下:
因为cos^2x+sin2x=1,所以cos^2x=1-sin^2x
方程可化为1-2 sin^2x=1/2,所以sinx=1/2.或-1/2.
所以在区间[-2π,2π]上,x=11π/6、7π/6、5π/6、π/6、-π/6、-5π/6、-7π/6、-11π/6