已知向量
(1)若x∈[2π,3π],求函数f(x)的单调递增区间; (2)若x∈(-
|
(1)f(x)=
•a
=cosx(cosx-3)+sinx(sinx-3)=1-3b
sin(x+2
),由 2kπ-π 4
≤x+π 2
≤2kπ+π 4
,k∈z,π 2
可得 2kπ-
≤x≤2kπ+3π 4
,再由 2π≤x≤3π 可得,2π≤x≤π 4
,9π 4
故单调递增区间是[2π,
].9π 4
(2)由f(x)=-1 可得 1-3
sin(x+2
)=-1,可得sin(x+π 4
)=π 4
,∵x∈(-2 3
,π 4
),π 4
∴0<x+
<π 4
,∴cos(x+π 2
)=π 4
,tan2x=7 3
=sin2x cos2x
=-cos2(x+
)π 4 sin2(x+
)π 4 -[1-2sin2(x+
)]π 4 2sin(x+
)cos(x+π 4
)π 4
=
=-[1-2×
]2 9 2×
×2 3 7 3
.-5 14 28