Solving inequalities, about trigonometric functions, thanks!

This question is a bit bluffing.

It is not difficult to solve when you note that the maximum values of [sin(x)+cos(x)-1] and [sin(2x)] are both x= 4.

Solution: Let t=sinx+cosx

0≤x≤π/2

1≤t≤√2

then sin2x=2sinxcosx=(sinx+cosx) -1=t -1

The original inequality becomes.

2a(1-a)(t-1)+a(t-1) 3 shifts, separates, simplifys.

a t +2a(1-a)t+a -2a-3 0 let f(t)=a t +2a(1-a)t+a -2a-3=(at+3-a)(at-1-a).

Under condition 1 t 2, f(t) 0 is constant.

Obviously, when a=0, f(t) 0 is constant;

When a≠0, let f(t)=0, and get the two roots of the equation, t1=1-3 a, t2=1+1 a

1) When a>0, t2 > t1, to make f(t) 0 constant, just need.

t1≤1t2≥√2

The solution, 0t2, to make f(t) 0 constant, it is only necessary.

t2≤1t1≥√2

Solution, -3(1+ 2) a<0

In summary, the value range of a is -3(1+ 2) a 1+ 2

Fundamental inequality.

The arithmetic square root is greater than or equal to the geometric square root. Note that both a and b must be greater than or equal to 0.

sinx=1 2, must be in the first and second quadrants.

x=6+2k, or x=5 6+2k, in the first quadrant, increments.

In the second quadrant, decreasing.

So 6+2k x 5 6+2k

then x=2k + 6, or x=2k +5 6,

（2kπ,2kπ+π2)

Composed. cosx 0 derives that it should be (2k - 2, 2k + 2).

To take advantage of the periodicity of trigonometric functions, you can generally look at it in one cycle.

The title is wrong. It should actually be 2, as evidenced by the following:

x²-2x-1=(x-1)²+2≥2.

Because all three numbers are positive, it is preferred to take the minimum value when the three numbers are equal.

Looking back, this topic is almost verbal.

a＝b＝π/4，c＝π/2

Segmentation, remove the absolute value number.

At x<3: Left side of inequality =-(x-5)-(x-3)=8-2x3 x 5: Left side of inequality =-(x-5)+(x-3)=2x>5:

Left side of inequality = +(x-5)+(x-3)=2x-8 Obviously there is 2 to the left of the inequality so when a 2, there is no solution to the inequality.

(x+3) (x-1) ≠ 0, x≠-3 or 1

If -3 x 1 then (2x-1)(x+1) ,1[ [1 2,+1)If x 3 1, then (2x-1)(x+1) 0, no solution to the sum up, x -3,-1] [1 2,1).

The third is to square the two sides to remove the absolute value and then solve it.

Three flowers on the same number line, three common parts are.