The known function f x sin 2x 6 cos x

Solution: f(x)=cos x-(3)cosxsinx+1

f(x)=2[cos²x-(√3)cosxsinx+1]/2

f(x)=[2cos²x-2(√3)cosxsinx+2]/2

f(x)=[2cos²x-1-(√3)·2cosxsinx+3]/2

f(x)=[cos(2x)-(3)sin(2x)+3]/2

f(x)=(1/2)cos(2x)-[3)/2]sin(2x)+3/2

f(x)=sin(π/6)cos(2x)-cos(π/6)sin(2x)+3/2

f(x)=sin(π/6-2x)+3/2

f(x)=3/2-sin(2x-π/6)

The monotonically increasing interval of f(x) is: 2k - 2 2x - 6 2k + 2, where: k = 0, 1, 2, 3 ......, the same below.

2kπ-π/3≤2x≤2kπ+2π/3

kπ-π/6≤x≤kπ+π/3

That is, the monotonically increasing interval is found as: x [k - 6, k + 3].

With f(x)=3 2-sin(2x- 6) found above, the second question is simple.

Leave it to the landlord to practice.

sin2x = three-thirds of the root number.

So sinxcosx = the root of sixths number three. Beg.

f(x)=cos x-( beat3)cosxsinx+1

f(x)=2[cos²x-(√3)cosxsinx+1]/2

f(x)=[2cos²x-2(√3)cosxsinx+2]/2

f(x)=[2cos²x-1-(√3)·2cosxsinx+3]/2

f(x)=[cos(2x)-(3)sin(2x)+3]/2

f(x)=(1/2)cos(2x)-[3)/2]sin(2x)+3/2

f(x)=sin(π/6)cos(2x)-cos(π/6)sin(2x)+3/2

f(x)=sin(π/6-2x)+3/2

f(x)=3/2-sin(2x-π/6)

The monotonically increasing interval of f(x) is: 2k - 2 2x - 6 2k + 2, where: k = 0, 1, 2, 3 ......, the same below.

2kπ-π3≤2x≤2kπ+2π/3

K - 6 x Brother Raider K + 3

That is, the monotonically increasing interval is found as: x [k - 6, k + 3].

With f(x)=3 2-sin(2x- 6) found above, the second question is simple.

Leave it to the landlord to practice, 2, sin2x = three-thirds of the root number.

So sinxcosx = the root of sixths number three.

Find, 0, known function f(x) sin(2x- 6)+cos first dig x

If f(x) 1, find the value of sinx*cosx.

Solution: (1) Using the difference angle formula and the double angle formula to simplify the function, f(x) = 3 2 sin2x + 1 2 can be obtained, and from f( )1, the mu book sin2 3 3 can be obtained, so that it can be found;

2) According to the monotonicity of the sinusoidal function, the monotonic interval of the function 1) f(x) sin2xcos can be obtained

6−cos2xsin

1+cos2x

2sin2x+

From f( )1, sin2 is obtained

3, so sin cos

2sin2θ＝

2) When 2+2k 2x

2+2k, k z, i.e. x [

4+k,4+k],k z,f(x) monotonically increases So, the monotonically increasing interval of the function f(x) is [

4+k, 4+k],k z, monotonically reduced interval for (4+k, 3

4+kπ)，k∈z．

1. The known function f(x) sin(2x 6 )+co s 2 x

1) If f( )1, find the value of sin cos;

2) Find the monotonic interval of the function f(x).

f(x)=√3sinxcosx-cos²x+1/2=√3/2*sin2x-1/2(2cos²x-1)=√3/2*sin2x-1/2*cos2x=sin(2x-π/6)

1) Minimum positive period: t=2 2=

Let sin(2x-6) = 1

Then 2x- 6= 2+k, k z

The axis of symmetry equation is: x= 3+k2, kz(2), in abc.

f(a2)=sin(a-6)=1 2 a-6= 6 or 5 6

i.e. a= 3 or (rounded).

cosa=(b +c -a) 2bc=1 2 and bc=6

a²=b²+c²-6

2BC-66 then the minimum value of a is: 6

f'(x)=1+2sinxcosx+cos2x-1=√2sin(2x+π/4)

1) The monotonic increase interval is 2k - 2=<2x+ 4<=2k + 2, i.e. k -3 8=2) 3 4=<2x+ 4<=3 2+ 4, the maximum value x= 4, f(x)=1

The minimum value x = 5 8, f(x) = - 2

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