Question

.若f(1+sinX)=cos2X-3sinX,则f（x）=？

Answer 1

f(1+sinX)=cos2X-3sinX
=1-2(sinx)^2-3sinx
=-[2(sinx)^2+3sinx-2]-1
=-(2sinx-1)(sinx+2)-1
=-[2(1+sinx)-3][(1+sinx)+1]-1

f(x)=-(2x-3)(x+1)-1=-2x^2-x+2

Answer 2

令a=sinx
则
cos2x-3sinx=1-2sin²x-3sinx=1-2a²-3a
所以f(1+a)=-2a²-3a+1
令x=1+a
a=x-1
f(x)=-2(x-1)²-3(x-1)+1
即f(x)=-2x²+x+2