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With the double angle formula:

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

Then use the properties of the quadratic function.

Let t=cosx, then -1<=t<=1

That is, find the minimum value of y=-2t 2+t+1.

The axis of symmetry of this quadratic function is t=1 4

So it increases monotonically on [-1,1 4] and decreases monotonically on [1 4,1], so just try t=-1 and t=1 to see which is smaller. When substituting finds t=-1, y=-2 is the smallest.

So the minimum value of y is -2

y=cosx-cos2x=cosx-(2cosx^2-1)=1+cosx-2cosx^2

cosx=1 4. The function is maximum, y=1+1 4-1 8=9 8

When coxx=-1, the minimum function is y=1-1-2=-2

y=cosx-cos2x=cosx-(2cosx^2-1)=1+cosx-2cos^2

The image of this function is symmetrical with respect to x=1 4, where x is minus one and ymin is minus two.

y=cosx-2cos²x+1

cosx belongs to [-1,1].

When cosx=-1, y obtains a minimum value of -2

Because it is square + square + absolute value = 0

squared, the absolute value must be greater than or equal to 0

So here it is only possible:

a-2b-4=0

2b+c=0

a-4b+c=0

Three unknowns, three equations, and you can solve it.

a-2b-4=2b+c=0

So a-4b-4-c=0=a-4b+c

c=-2 is substituted into 2b+c=0

Get b = 1 and substitute a-2b-4 = 0

Get a=6 because a=6, b=1, c=-2

So substituting 3a+b-c gets 21

Because (a-2b-4) +2b+c) =-|a-4b+c|, so (a-2b-4) +2b+c) =0, a-4b+c=0

This gives a-2b-4=0

2b+c=0

a-4b+c=0

Finally, we find a=6 b=1 c=-2

（a-2b-4)²+2b+c)²+a-4b+c|=0 to obtain: a-2b-4=0;2b+c=0；a-4b+c=0 is obtained by the third formula: a-2b-4b+2b+c=0 Substituting the first and second formulas into the first formula yields b=1, c=-2, a=6

then 3a+b-c=21

（a-2b-4)²+2b+c)²+a-4b+c|=0 because: if the sum of several polynomials is zero, then each term of this polynomial is zero, so: a-2b-4=0 , 2b+c=0 , a-4b+c=0 , expressed by the algebraic expression of b, and then substituted into obtain:

a=6，b=1，c=-2

Substituting a=6, b=1, c=-2 into 3a+b-c yields: 3 6+1-(-2)=21

The root of the equation a = root number 3, b = 2 - root number 3

c=2, because a+b=c, the triangle does not exist.

The first equation has no solution to the problem.

Solution: x -2x + 3 (2-3) = 0

x1=√3 ； x2=2-√3

a=√3，b=2-√3

x²-4=0

x1=2 ;x2=-2

c=2 and a+b=c

Triangles with a, b, and c as sides do not exist.