The known functions f x 2sin pie x 8 pie 4 1

Substituting x into f(x) to get the equation of x, and then g(x) will know the intersection of the x-axis, then y is equal to o, that is, g(x)=0, and the next thing is to simplify the sin bracket and open it, there is a formula, this will not happen, you don't have to do ......Two will be eliminated, and then there will be two sinxcos....Addition Merge together, there is also a formula, do it yourself, you do it yourself, and see this is basically what you do.

sin(2x+pie 6)=3 cover guess 5, 2x+pie 6=2k +37° or 2k +143°, the width of the object is x1=k +3.5° or k +56.5° (where pie 6=30°, in general, two right-angled sides are 3 and 4 right-angled trigram angles, and its two acute angles are about 37° and 53°If x1 belongs to [faction 4, faction 2], then x1 = 56 points 5°

Solution: (1)f(x)=2sin (4+x)- 3cos2x=1-cos(2+2x)- 3cos2x=1+sin(2x)- 3cos2x

1+sin(2x-π/3)/2

The maximum and minimum values of 4 x 2, 2 2x , 6 2x- 3 2 3,1 4 sin(2x- 3) 2 1 2,5 4 f(x) 3 are 3 2 and 5 4, respectively.

2) If f(x) m, then there is f(x)-m<2,5 4 f(x)-3 4, combined with f(x) m, there is -3 4 if f(x).

Solution: g(x)=2msin(2x+ 6)+n because x[0, 2].

So 2x [0, ].

2x+π/6∈[π6,7π/6]

When 2x+6=2, the maximum value of 2m+n=1 is obtained, but when 2x+6=7 6, the minimum value of -m+n=-5 is obtained by - to obtain 3m=6

m = 2 substitution to get n = 1-2 m = 1 - 4 = -3

So m=2n=-3

1): The function f(x)=2sin(x+ 3)-2sinx=3cosx-sinx, x belongs to [- 2,0]; cosx = 3 3, x belongs to [- 2,0]; sinx=

f(x)=√3cosx-sinx=1+(√6/3)；

2): f(x)=2sin(x+ 3)-2sinx= 3cosx-sinx=2cos(x+ 6);

x belongs to [- 2,0];

x+6 belongs to [- 3, 6];

cos(x+6) belongs to [1, 2,1];

The value range of f(x)=2cos(x+ 6) is:[1,2];

f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)

3sin(x+π/2)+sinx

3cosx+sinx

2sin(x+ π/3),1)2π.

2) g(x)=2sin(x+6), the maximum value on [0, ] is 2, and the minimum value is -1

(1):

f(x)=sin(x)+cos(x)) = √2(sinx*√2/2+cosx*√2/2) = √2sin(x+π/4)

So f(x)=2f(x+ 2) = 2 2sin(x+ 2+ 4) = 2 2cos( 4+x).

i.e. 2sin(x+ 4) = 2 2cos( 4+x) i.e. tan(x+ 4) = 2

2)g(x) = f(x)f(x+π/2) = √2sin(x+π/4) *2sin(x+π/4+π/2)

2sin(x+π/4)cos(x+π/4)=sin(2x+π/2)=cos(2x)

Minimum: -1 The set of corresponding x.

Solved.

f(x)= 3sin x +cos x=2(sin xcos 6+cos xsin Zheng 6)=2sin( x+ 6).

1≤sin(πx+π/6)≤1

f(x) maximum value 2, minimum value -2

The minimum positive period t = 2 2

Monotonically increasing interval regression: 2k - 2 x + 6 2k + 2, i.e. x (2k-2 3, 2k+1 3), where k z;

Motonic reduction interval: 2k +2 2 x+ 6 shout Shiqiao 2k +3 2, that is, x (2k+1 3,2k+4 3), where k z;