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Hello, the following is my thinking and solution, please refer to it.

Since 2cosx = (1+sinx)- 1-sinx), the equation still holds after the two sides of the equal sign are squared at the same time, i.e., (2cosx) 2=( (1+sinx)- 1-sinx)) 2.

The key is that the calculation on the right side uses the squared difference and the perfect square formula, and the right side = (1+sinx)+(1-sinx)-2( (1+sinx)(1-sinx)).

2+2(√1-(sinx)^2)=2+2(√(sinx)^2+(cosx)^2-(sinx)^2)=2+2(√(cosx)^2)

Next, we will discuss the problem of absolute values separately, i.e., the original form can be divided into the following two cases:

2(cosx)^2-cosx-1=0 (1)

2(cosx)^2+cosx-1=0 (2)

Equations 1 and 2 are deformed, respectively.

2cosx+1)(cosx-1)=0 (1')

2cosx-1)(cosx+1)=0 (2')

By (1') to obtain cosx=-1 2 or cosx=1

Thus x=2k - 3 or x=2k (k z);

By (2') to obtain cosx=-1 2 or cosx=1

therefore x=2k + 3 or x=(2k+1) k z);

Therefore, the synthesis (1') and (2'x=2k 3 or x=k (k z);

This makes it easy to find:

When x=2k3 (k z), tagx= 3;

When x=k (k z), tagx=0.

The above is my thinking and solution, please refer to it.

2cosx = (1+sinx)- 1-sinx) squared, 4cos 2(x) = 1+sinx+1-sinx-2 root number (1-(sinx) 2).

4cos 2 (x) = 2-2 root number (cos 2 (x)) 2cos 2 (x) + cosx-1 = 0 (cosx>0) cosx + 1) (2cosx-1) = 0

cosx=-1 (rounded) cosx=1 2

sinx=+ - root number 3 2 (negative value rounded).

Because it does not comply with cosx>0

When cosx<0

Yes: cosx=-1 2

sinx = - root number 3 2

tanx=sinx cosx root number 3

√（1+sinx）-√1-sinx）][1+sinx）+√1-sinx）]=2sinx

1+sinx）+√1-sinx）=sinx/cosx（1+sinx)=(tanx+2cosx)/24+4sinx=tan^2+4cosx^2+4sinxcosx^2=1/4

In both cases, tanx is the root number 3

Square on both sides. There are 4cos x=1+sinx+1-sinx-2 1-sinx

1-sinx²=cosx²

4cos x=2 2cosx, where cosx is positive.

2cos²x+cosx-1=0

2cosx-1)（cosx+1)=0

cosx=1/2

sinx=√3／2

tanx=sinx／cosx=√3

First of all, the square of the two sides: 4cos 2x = 2-2 root number (1-sin 2x), because 1-sin x = cos 2x, simplified to 4cos 2x = 2-2 root number cos 2xThen simplify it to get 2cos 2x+|cosx|-1=0

Solution|cosx|=1/2.So tanx = root number 3

The square of both sides gets: 2*t 2=1-t. where t=|cosx|〉=0，t=1/2。Then use the induction formula or the universal formula.

1. x=12, collision collapse a=b

3、x=2/11

4. ab>rotten hall 0

5. Equal. "Laughing Circle x<2

The formula method calculates that the value of x is with a, and then one of the two values is greater than 2 and less than 2, and the range of a is determined.

There are two causes: 4a, 2-4>0

a<-1 or a>1

Let x1 and x2 be the two roots.

(x1-2)(x2-2)<0

x1x2-2(x1+x2)+4<0

1-2*(-2a)+4<0

a<-5/4

So a<-5 4

This question can be simply calculated like this.

f(2)<0 therefore a <

Hope it helps.

The hyperbolic properties give PF PF =2a

m is the midpoint of the line segment pf, and of = of

mf₁=½pf₁mo=½pf₂

Yes |mf₁|-mo|= a=3

The line segment f t is the tangent of the circle.

F to is a right-angled triangle.

f₁t²+ot²=f₁o²

Gotta |f1t| =5

MO is the median line of the triangle F10F2, and according to the definition of hyperbola, it can be known that MF1-MO is equal to 3

The triangle F10T is a right triangle, so F1T can be calculated as 5

｜pf₁｜－pf₂｜=2a

m is the midpoint of the line segment pf, and of = of

mf₁=½pf₁mo=½pf₂

Yes |mf₁|-mo|= a=3 f to is a right triangle f t +ot =f o ot=3, f o= (9+25) f1t| =5

a= b= a∩b=？

y=(x-3) -4 x r in a, then y -4, then y=(x-1) +2 x r in a=b, then y 2, i.e. b = because 2>-4

then a b=a= b= and a b=b then a=?

If the equation in a is solved x=6 or x=-1, then a=

If a=0 in b, then ax+1=0 becomes 1=0, obviously there is no solution, and b is an empty set, satisfying a b=b

If a≠0, then x=-1 a, i.e. b=

If a b = b

then -1 a = 6 or -1 a = -1

then a=-1 6, or a=1

In summary, a=0, -1, 6 or 1

a= b= a b=b then what is the range of values for a?

a b = b requires b to contain a completely

then a<2

a=y=x 2-6x+5=(x-3) 2-4 -4 therefore a=y=-x 2-2x+3=(x-1) 2+2 2 therefore b= a b=bx 2-5x-6=0 (x-6)(x+1)=0 x=-1 x=6 therefore a

ax+1=0 If a=0 has no solution, b is an empty set that satisfies the condition.

If a≠0 b= a b=b then -1 a=-1 or -1 a=6 gives a=1 or a=-1 6

In summary, a=1 or a=-1 6 or a=0

a b = b indicates that a is a subset of b so the scope of b should include the scope of a.

That is, a<2 (you can't take the equal sign, if you take the equal sign, there is an element of 2 in a, but there is no one in b, and the condition is not met).

The answer is a<2

1.a, y=x 2-6x+5=(y-3) 2-4, i.e. y>=-4; b,y=-x 2-2x+3=(y-1) 2+2, i.e. y>=2.So a b=b=

2.a= , solve the equation x=6 or x=-1If a=0, unsolved b is an empty set, if a≠0, b= gives a=-1 x, then a=-1 6, or 1So a=1 or a=-1 6 or a=0

3.It is relatively simple and does not require process a<2

Crossing point B to do BF is perpendicular to AO and point F

Passing point C to do CE perpendicular to point Bo to point E

Let be=x, of=y, and because c=60°, then ce=root number 3x, bo=2y, be=2y-x, fc=y-2x

So we can get the square of the system of equations (2y-x) + the square of the root number 3x = the square of 21.

31 squared - [20 - (y-2x)] squared = 21 squared - (y-2x) squared.

Solve x and y.