(2013•南开区一模)已知函数f(x)=sin(2x+π6)−cos(2x+π3)+2cos2x.
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(2013•南开区一模)已知函数f(x)=sin(2x+
)−cos(2x+
)+2cos
π |
6 |
π |
3 |
(1)f(
π
12)=sin(2×
π
12+
π
6)−cos(2×
π
12+
π
3)+2cos2
π
12=sin
π
3−cos
π
2+1+cos
π
6=
3
2−0+1+
3
2
=
3+1
(2)∵f(x)=sin(2x+
π
6)−cos(2x+
π
3)+2cos2x
=sin2xcos
π
6+cos2xsin
π
6−cos2xcos
π
3+sin2xsin
π
3+cos2x+1
=
3sin2x+cos2x+1=2sin(2x+
π
6)+1,
∴当sin(2x+
π
6)=1时,f(x)max=2+1=3,
此时,2x+
π
6=2kπ+
π
2,即x=kπ+
π
12)=sin(2×
π
12+
π
6)−cos(2×
π
12+
π
3)+2cos2
π
12=sin
π
3−cos
π
2+1+cos
π
6=
3
2−0+1+
3
2
=
3+1
(2)∵f(x)=sin(2x+
π
6)−cos(2x+
π
3)+2cos2x
=sin2xcos
π
6+cos2xsin
π
6−cos2xcos
π
3+sin2xsin
π
3+cos2x+1
=
3sin2x+cos2x+1=2sin(2x+
π
6)+1,
∴当sin(2x+
π
6)=1时,f(x)max=2+1=3,
此时,2x+
π
6=2kπ+
π
2,即x=kπ+
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