Question

設(shè)α∈R,函數(shù)f(x)=sin 2xcos α+cos 2xsin α-cos(2x+α)+cos α,x∈R.

(1)若α∈[,],求f(x)在區(qū)間[0,]上的最大值;

(2)若f(x)=3,求α與x的值.

Answer 1

解:(1)易知f(x)=sin(2x+α)-cos(2x+α)+cos α

=2sin(2x+α-)+cos α,

由于α-∈[0,],

2x+α-∈[α-,α+],

所以當(dāng)2x+α-=,

即x=-時(shí),f(x)_max=2+cos α.

又f(x)_max=2+cos α在α∈[,]上單調(diào)遞減,

所以f(x)_max=2+cos α≤2+,

當(dāng)α=時(shí)取到最大值.

綜上可知,當(dāng)α=,x=時(shí),f(x)_max=2+.

(2)由于f(x)=2sin(2x+α-)+cos α,

且2sin(2x+α-)≤2,cos α≤1,

現(xiàn)在已知f(x)=3,則等價(jià)于

解得