High School Math Trigonometry. Process. Question 5.

1.When a=0, it is a one-dimensional equation, and x=-1 2 satisfies the requirements of the problem.

2.When a is not 0, it is a quadratic equation and must have a real root.

x1 + x2 = -2/a

x1*x2 =1/a

b^2-4ac = 4-4a>=0

Solution: a<=1, and a is not 0

1) If x1 and x2 are both negative, then x1 + x2 = -2 a < 0 and x1*x2 =1 a >0 solve: 0 then x1*x2 =1 a <0 solves: a<0 In summary, a<= 1, the problem requirements are satisfied.

Because there must be a solid root, then a≠0 when 4 4a 0, a 1, the intersection of the image and the x-axis is 1,0 , which is true.

When 4 4a 0, a 1, when 1 a 0, the abscissa of the intersection point to the left of the intersection point with the x-axis is b δ 2a, according to the title, b δ 2a 0, 2a 0, b δ 0, δ0, b δ, holds.

When a 0, the abscissa of the intersection point to the left of the intersection point of the x-axis is b δ 2a, and according to the title, b δ 2a 0, 2a 0, b δ 0, δ0, δb, holds.

When 4 4a 0 and a 1, the graph and the coordinate axis have no intersection point.

In summary, 0 a 1 or a 0

Sub-situation: 1When a=0, it is a one-dimensional equation, and x=-1 2 satisfies the requirements of the problem.

2.When a is not 0, it is a quadratic equation and must have a real root.

x1 + x2 = -2/a

x1*x2 =1/a

b^2-4ac = 4-4a>=0

Solution: a<=1, and a is not 0

1) If x1 and x2 are both negative, then x1 + x2 = -2 a < 0 and x1*x2 =1 a >0 solve: 0 In summary, a<= 1, the problem requirements are satisfied.

So, obviously, this answer of yours is wrong, dear

If you still have doubts, please feel free to ask and ask.

Solve using Vedder's theorem and discriminants.

Because sin2a = sin2b angle a is either equal to angle b or equal to (180 degrees - angle 2a) e.g. sin30 degrees = sin150 degrees.

Assuming that angle a is equal to angle b, this situation does not exist. Because there is already 0 above that is less than a and less than b than 180 degrees, it is assumed that it does not exist.

So angle 2b can only be equal to (180 degrees - angle 2a).

I think the answer is very detailed, is there any problem?

a-b=(sinx-cosx,o)

A times (a-b) = (sinx-cosx, 0) times (sinx, cosx) oak shed = sinx 2-sinxcosx = 1-cosx 2-1 and manuscript 2sin2x

1 2-1 2 (sin2x + cos2x) = 1 2 - 2 times the root number of the second half of the sin (2x + the party of the four beams).

Therefore, the minimum positive period is pie.

2sinxcosx=sin2x

(1) Derived from the original formula.

f(x)=－√3sin²x+

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

Substituting x=25 6 gives the value as sin(2 3+8 )-3 2=0

2) f(a 2) = sin(a + 3) - 3 2 = 1 4- 3 2, a (0, ) then sin(a + 3) = 1 4<1 2, cos(a + 3) = - 15 4, then sina = sin(a + 3 - 3) = sin(a + 3) * (1 2) - cos(a + 3) * (3 2).

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

The first question is f(25 6) = f( 6) = sin(2 3) - 3 2 = 0

The second question is f(a 2) = sin(a + 3) - 3 2 = 1 4- 3 2 sin (a + 3) = 1 4

1/2sina+√3/2cosa=1/4

There is sin 2a + cos 2a = 1

So sina = (1 + 3 5) 8

（1）f(x)=－√3sin²x+

3 2*cos2x+1 2*sin2x- 3 2=sin(2x+ 3)- 3 2 (1) so f(25 6)=sin(26 3)- 3 2=sin(2 3)- 3 2=0

2) Because f(a 2) = 1 4- (3) 2 is substituted for (1), sin(a+ 3) = 1 4

Thus sina * cos 3 + cosa*sin 3 = 1 4 and because sina 2 + cosa 2 = 1 from a (0, ) finally get sina = (1 + 3 5) 8

Let t=sinx, x [0, 2].

t∈[0，1]

The function becomes: y=t +t 2

y=t +t2 increases monotonically on [0,1], and ymin=0 when t=0

t=1, ymax=3 2

The value range is [0,3 2].