When x 6, 7 6 the function y 3 sinx 2cos squared 2 x range

y=3-sinx-2(1-(sinx)^2)2(sinx)^2-sinx+1

2(sinx-1/4)^2+7/8

Because x [ 6,7 6] so sinx [-1 2,1] so (sinx-1 4) 2 [0,9 16] so y=2(sinx-1 4) 2+7 8 [7 8,2] generally such a problem is either followed by the square of cos, and the square relation is replaced by the square of sin, and the formula is completed; Either cos2x is lowered and raised with a double angle formula, and then the recipe is completed. It can also be completed without the processing method of converting the formula into a quadratic function with a fixed axis and a fixed interval. I hope the title is correct.

Functionalize as y=2sin x squared --sin x +1=2 (sin x --1 4) squared +7 8

2.Let sin x=t, then y=2(t-1 4) squared +7 8, and v6"x"7 vultures 6 can get -1 2"t"1

3.So when t=1 4, y has a minimum value, y(min)=7 8; When t=-1 2 or t=1, y has a maximum value and y(max)=2

4.In summary, 7 8 "y" 2, that is, the value range is [7 8,2].

The function y=3-sinx-2cos x,x [ 6,7 6] then sinx [-1 2,1].

y=3-sinx-2(1-sin x)=2sin x-sinx+1 is a quadratic function, and with respect to the quadratic function of sinx, there is a minimum value of 7 8 when the axis of symmetry sinx=1 4, and a maximum value of 2 when sinx=1

So the range is [7, 8, 2].

When x [ 6,7 6], sinx [1 2,1],y = 3-sinx-2(1-sin x) = 2sin x-sinx+1, it can be seen from the nature of the parabola that this quadratic function increases monotonically on [1 2, 1], so y [1, 2].

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x [-12,5 12 ], then 2x-6 has a range of [-3,2 3] and sin(2x-6) has a range of [-1 2,1].

So the range of the function y is [,3].

y=3-sinx-2cos²x

3-sinx-2(1-sin²x)

2sin²x-sinx+1

2(sinx-1/4)²+7/8

This is about the "quadratic function" of sinx

x∈[π6,7π/6]

sinx∈[-1/2,1]

When Duan Zheng sinx = 1 hall 4, y obtains the minimum value of 7 8 and when sinx = 1 or 1, y obtains the maximum value of the pretend burning macro 2

The function y=3-sinx-2cos x, brother letter x [ 6,7 6] then sinx [-1 2,1].

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

Quadratic functions. About the quadratic function of sinx, the axis of symmetry or the sock.

sinx=1 4 has a minimum value of 7 8, sinx=1 has a maximum value of 2 envy round so.

Range. Yes [7 8, 2].

2(1-sin²x)+sinx-1

2sin²x+sinx+1

2(sinx-1/4)²+9/8

x∈（-6，π/3）

sinx∈(-1/2,√3/2)

While. sinx=1 4.

There is a maximum value for.

When the tomb is opened. sinx=-1 2.

There is a minimum value for.

The known Luwang Bridge is based on the value range.

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 the sin beam is taken as sinx=1 2, y has a minimum value of 5 2

When sinx=1, there is a maximum value of 5, and the value slag circle field is [5 2,5].

y=3-sinx-2cos²x

2（sinx-1/4）²+7/8

When x [ 6,7 6] , the function y=3-sinx-2cos x has a range of [7 8,2].

y=4sin^2x+6cosx-6

4-4cos^2x+6cosx-6

4cos^2x+6cosx-2

Let a=cosx

Negative thirds are less than or equal to x and less than or equal to two-thirds.

Then -1 2<=a<=1

y=-4a^2+6a-2

4(a-3/4)^2+1/4

So when a = 3 4, y is maximum = 1 4

A = -1 2, y minimum = -6

value range [-6,1 4].

Solution: Let the front f(x)=2cos x+5sinx-4 then f(x)=2(1-sin x)+5sinx-4-2sin x+5sinx-2

2 (sinx-5 Sun Gaosheng 4) +9 8

Known by (3 x Nian Da 5 6).

1/2≤sinx≤1

So f(x)min=f(1 2)=0

f(x)max=f(1)=1

Hence the range f(x) [0,1].