It is known that the minimum positive period of the function f x 3sin x 2sin 2 x 2 0 is . If f x 2 1

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

2sin(ωx+π/6)

So the minimum positive period of f(x) t=2=, i.e. =2, f(x)= 2sin(2x+ 6)

f(x 2) = 1 3, x ( 2, ) so f(x 2) = 2sin(x + 6) = 1 3, i.e. sin(x+ 6) = 1 6, because x ( 2, ) so x+ 6 (2 3, 7 6), cos(x+ 6) = - 35 6

sinx=sin[(x+π/6)-π/6]=sin(x+π/6)cosπ/6-cos(x+π/6)sinπ/6=1/6*√3/2-(-35/6)*1/2

Original formula=3sin x-(1-cos x)=3sin x-cos x-1=asin( x-k)-1, so =2;Bring in the original formula to get, f(x)= 3sin2x-cos2x-1 and substitute f(x 2)=1 3 to get 3sinx-cosx-1=1 3! Simplification: 3sinx-cosx=4 3

Coupled with cosx 2+sinx 2=1, 10sinx 2-8sinx-7 9=0; Then use the root finding formula to get two results, and round off one according to the range of x!

f(x)=2sin^2 (x+π/4)-√3cos2x-1=-cos(2x+π/2)-√3cos2x=sin2x-√3cos2x

2sin(2x-π/3)

When x belongs to r, the minimum positive period of the function f(x) t=2 2=

To reduce the power expansion angle, first convert the square of the sine into a cosine, and then use the induction formula to convert it into a sinusoid, and in the third step, use the auxiliary angle formula to convert it into the form of y=asin(wx+%), the specific process:

x)=2sin^2 (x+π/4)-√3cos2x-1=-cos(2x+π/2)-√3cos2x=sin2x-√3cos2x

2sin(2x-π/3)

When x belongs to r, the minimum positive period of the function f(x) t=2 2=

(1) Analysis: The minimum positive period of the function f(x)=sin 2wx+ 3sinwxsin( 2+wx) (w>0) is

f(x)=sin^2wx+√3sinwxsin(π/2+wx)=1/2-1/2cos2wx+√3/2sin2wx

sin(2wx-π/6)+1/2

t=π==>2w=2==>w=1

f(x)=sin(2x- 6)+1 2 monotonically increasing interval: 2k - 2<=2x- 6<=2k + 2==>k - 6<=x<=k + 3

2) Parsing: on the interval [0,2 3].

f(0)=sin(-π/6)+1/2=0

f( 3)=sin(2 3- 6)+1 2=3 2f(2 3)=sin(4 3- 6)+1 2=0 f(x) in the range [0, 3, 2] on the interval [0, 3 3].

f(x)=2sinx/4cosx/4-2√3sin²x/4+√3=sin(x/2)+√3[1-2sin²(x/4)]=sin(x/2)+√3cos(x/2)

2[1/2*sin(x/2)+√3/2*cos(x/2)]=2sin(x/2+π/3)

The minimum positive period t=2 (1 2)=4

When x 2 + 3 = 2k + 2, k z

That is, when x = 4k + 3, k z, f(x) obtains the maximum value of 2 when x 2 + 3 = 2k - 2, k z

That is, when x=4k -5 3, k z, f(x) obtains a maximum value of -2

Minimum positive period = 2 2 = 2 = 2

1 When 4x- 6=k + 2, i.e. x=k, 4 + 6 is the axis of symmetry.

2 When 4x- 6 [k - 2, k + 2] the function monotonically increases k - 2<=4x- 6<=k + 2 k 4- 12<=x<=k 4+ 6

That is, the monotonic increase interval is [k 4- 12,k 4+ 6]3, when x [-12, 2] 4x- 6 [-2,11 6], so the maximum value is 1-1 2=1 2, and the minimum value is -1-1 2=-3 2, so the value range is [-3 2,1 2].

Solution: 2 = 2 t = 4

So =2, so f(x)=sin(4x- 6)-1 2(1)From 4x- 6= 2+k, x= 6+k, 4(2)- 12+k 2 < = 4x- 6< = 2+2k - 12+k 2< = x< = 6+k 2

So the monotonic increase interval is [k 2- 12, k 2 + 6] (3).From -12<=x<=2 - 2<=4x- 6<=11 6

So -1<=sin(4x-6)<=1

So -3 2<=f(x)<=1 2

So the range is [-3 2, 1 2].

The minimum positive period t=2 w=2

w=4f(x)=sin(8x- 6)-1 2 The axis of symmetry equation is.

8x-π/6=kπ+π2

x=k 8 + 12 (k is an integer).

2) Let 8x- 6=t

Then the increase interval of t is [2k - 2, 2k + 2]x, and the increase interval is [k 4- 24, k 4 + 12] (k is an integer) (3) 2-(-12) > the minimum positive period 2, so the value range is the value range in one period.

fmax=1-1/2=1/2

fmin=-1-1/2=-3/2

i.e. [-3 2, 1 2].

f(x)=sin²(x+π/4)+cos²(x-π/4)-1=(sinx+cosx)²*2/2)²+cosx+sinx)²(2/2)²-1

sinx+cosx)²-1

2sinxcosx

sin2xf(-x)=sin(-2x)=-sin2x=-f(x)f(x) is an odd function.

The positive period is: 2 2=