若sin(x+3п)cos(x-п)=2a+1,求实数a的取值范围.
问题描述：
若sin(x+3п)cos(x-п)=2a+1,求实数a的取值范围.
答
sin(x+3π)cos(x-π)
=sin(x+3π)cos(π-x)
=(-sinx)(-cosx)
=sinxcosx
=(1/2)sin2x=2a+1
sin2x=2(2a+1)
根据正弦函数定义域
因-1≤sin2x≤1
则-1≤2(2a+1)≤1
-1/2≤2a+1≤1/2
-1-1/2≤2a≤1/2-1
-3/2≤2a≤-1/2
-3/4≤a≤-1/4
答
sin(x+3п)cos(x-п)
=(-sina)(cosa)
=sinacosa
=1/2*sin2a=2a+1
则-1≤2a+1≤1
-2≤2a≤0
-1≤a≤0
答
sin(x+3п)cos(x-п)=2a+1
(-sinx)(-cosx)=2a+1
sinxcosx=2a+1
sin(2x)=4a+2
-1≤sin(2x)≤1
-1≤4a+2≤1
-3/4≤a≤-1/4
a的取值范围为[-3/4,-1/4]
答
sin(x+3п)cos(x-п)=2a+1
-sinx*(-cosx)=2a+1
sinxcosx=2a+1
1/2*2sinxcosx=2a+1
-1-1/2-3/2-3/4