y=|cos2x|+|cosx|=|2cos²x-1|+|cosx|=|2|cosx|-²1|+|cosx|,
設(shè)|cosx|=t≥0,y=|2t²-1|+t,
①√2/2≤t≤1時(shí),y=2t²-1+t=2(t+1/4)²-9/8
√2/2≤t≤1時(shí)函數(shù)單調(diào)遞增,y∈[√2/2,2]
②0≤t≤√2/2時(shí),y=1-2t²+t=-2(t-1/4)²+9/8,
此時(shí)t=1/4時(shí),函數(shù)取得最大值9/8,
t=√2/2時(shí),函數(shù)取得最小值√2/2.所以y∈[√2/2,9/8]
綜合①②知,函數(shù)y=|cos2x|+|cosx|的值域?yàn)閇√2/2,2].