Question

急求:高一数学:若根号1-sinx/1+sinx=sinx-1/cosx,则x的取值范围是

Answer 1

√[(1-sinx)/(1+sinx)]=(sinx-1)/cosx,
将式中所有的1都换成sin²(x/2)+cos²(x/2),sinx都换成2sin(x/2)cos(x/2),cosx换成cos²(x/2)-sin²(x/2),得
√{[sin²(x/2)-2sin(x/2)cos(x/2)+cos²(x/2)]/[sin²(x/2)+2sin(x/2)cos(x/2)+cos²(x/2)]}
=
-[sin²(x/2)-2sin(x/2)cos(x/2)+cos²(x/2)]/[cos²(x/2)-sin²(x/2)],
√{[sin(x/2)-cos(x/2)]²/[sin(x/2)+cos(x/2)]²}=
[sin(x/2)-cos(x/2)]²/{[sin(x/2)-cos(x/2)]*[sin(x/2)+cos(x/2)]},
|
[sin(x/2)-cos(x/2)]/[sin(x/2)+cos(x/2)]
|=[sin(x/2)-cos(x/2)]/[sin(x/2)+cos(x/2)],
上式实际上是|a|=a,说明a>0,所以
[sin(x/2)-cos(x/2)]/[sin(x/2)+cos(x/2)]>0
分子分母相除大于0,所以二者同号,所以二者相乘也大于0,即
[sin(x/2)-cos(x/2)]*[sin(x/2)+cos(x/2)]>0,变形
[sin(x/2)]²-[cos(x/2)]²>0
-cosx>0,
(上面的等式是一个倍角公式)
cosx<0
所以
2kπ+π/2<x<2kπ+3π/2

Answer 2

先将原式化简若
1-sinx≠0
得1+sinx+cosx=0。提出根号2
得1+ √2sin(x+π/4)=0
所以√2sin(x+π/4)=-1所以sin(x+π/4)=-√2/2
即x+π/4=2kπ+5/4π或x+π/4=2kπ-1/4π
得x=2kπ+3π/2或x=2kπ+π
若
1-sinx=0则
显然成立此时x=2kπ+π/2
综上x=2kπ+π/2或x=2kπ+3π/2或x=2kπ+π
k∈Z