Let the function y 3 2sinx 2 3cosx, and find its value range data when I make it myself, and tell me

Hello! There is a very simple way to do this, as mentioned below, which is to think of it as a problem of the slope of a point and a point on a circle, as follows: the original function y= 2(3 2 --sinx) 3(2 3 --cosx) = 2 3 (3 2 --sinx) (2 3 --cosx) then:

3 2 --sinx) (2 3 --cosx) represents the slope of any point from point (3 2, 2 3) to the circle: x 2 + y 2 = 1. Draw a sketch and you will know that the slope ranges as:

3 4,15 4] So, the original function range is:[1 2,5 2]Note: This method should be relatively simple, I don't know whether the specific answer is right or not (the comparison has been many years), but the method is absolutely correct, I hope it can help you, if you don't understand, you can continue to ask, hope, thank you!

Bringing out 2 3 and becoming (3 2-sinx) (2 3-cosx) can be seen as the slope of the line connecting the point and the point (2 3, 3 2) on the unit circle. The value range can be calculated by combining numbers and shapes.

Using 2sinxcosx=(sinx+cosx) 2-1 overall commutation, it becomes a quadratic function.

The number you give is not special, you can only use the derivative to determine the station.

Solution: function y=-2sin(x 3-4).

Because -1 sin(x 3- scatter base refers to 4) 1, so -2 y 2

Therefore, the value range of the function is [-2,2].

The deformation obtains: 4ycosx-4y=2sinx+3, that is, 4ycosx-2sinx=3+4y, the left side uses the contraction hall formula to become the root number (4y 2+4) sin(x+ )3+4y, that is, the root number (4y 2+4) = sin(x+ )1,1], and then it is the solution to the inequality about y! Xiang Xiang summoned and you will solve it!

Hello! There can be a very simple way to solve this problem, which has also been mentioned below, that is, Pazhou regards it as a slope problem between a point and a point on the circle of Hemu, and the specific method is: the original function y=

sinx)/3(2/3

cosx)=

sinx)/(2/3

cosx) then: (3 2

sinx)/(2/3

cosx) is the slope of any point from point (3 2, 2 3) to a circle: x 2 + y 2 = 1.

If we draw a sketch, we know that the slope ranges as [3 4,15 4] so the range of the original function is: [1 2,5 2].

Note: This method should be relatively simple, I don't know if the specific answer is right or not (the comparison has been many years apart), but the method is absolutely correct.

Take the 2 3 Thyssen key out and become (3 2-sinx) This coincidence (2 3-cosx) can be seen as the slope of the line connecting the dot on the unit circle with the dot (2 3, 3 2). The value range can be calculated by combining numbers and shapes.

Solution: y=2-3(1-cos2x)-4cosx=3cos2x-4cosx-1, so that t=cosx, then y=3t2-4t-1,x [-3,2 3], 1 2 cosx 1, -1 2 t 1, the symmetry axis of the quadratic function y=f(t)=3t2-4t-1 is t=2 3, when t -1 2,1, this quadratic state source function is subtracted on t -1 2,2 3, and is an increase function on t 2 las3,1, ymin=f(2 3)=-7 3,ymax=f(-1 2)=7 4], i.e., the range of the function is -7 3,7 4].

Let's replace sinx with cosx

Make the answer stupid cosx=t

The t range is the source of locust [,1].

Original formula = 2-3 (1-t 2)-4t = 3t 2-4t-1 = 3 (t-2 3) 2-7 3

Monotonically decreasing on (,2 3) and increasing monotonically on (2,3,1).

The minimum value is x=2 3.

min=-7/3

Maximum=max(y(,y(1)))=max(1 4,-2)=1 4

Thank you very much! I don't know how to ask again.

Solution: y=-2(1-sin2x)+2sinx+3=2sin2x+2sinx+1

2(sinx+1 2)2+1 2, because x [ 6,5 6], so 1 2 sinx 1, so when sinx=1 2, y has a minimum value of 5 2 When sinx=1, there is a maximum value of 5, and the value range is [5 2,5].

y=1-2sinx+3cos x (utilizing cos 2x= 1- sin 2x).

y= -3sin 2x -2sinx +4, let t= sinx, then y=-3t 2 -2t +4, -1<=t<=1 This is a quadratic function problem.

The axis of symmetry x=-1 3 with the opening downward.

So y(min) = -3-2+4 =-1 (when t=1) y(max) = -3(-1 3) 2 -*1 3) 4 4 (when t 1 3).