If 0 sinx 1, find the range of x values

Solution: According to the problem, the value range of the graph is in the upper part of the abscissa x-axis: that is, the value of the positive rotation curve x is: 2k

2k where represents pi.

k is an integer.

Solution: Because of the sine function y

sinx and cosine function y

The minimum positive period of cosx is 2, so consider the case on x [-, plus 2k and k z is sufficient. A discussion of the classification of the value of x yields:

1) If cosx

0, i.e. x (- 2, 2), so sinxcosxtanx

x∈(-/2

kπ，π/4

k ), k z, so when k0, x

2) If cosx

0, sinx at this time

1, by the title sinx

1, at this time. x

3) If cosx

0, i.e. x (- 2, 2), so sinxcosxtanx

x∈(π/4

kπ，π/2

k ), k z, to make x (- 2) ( 2, ) so when k1, x In summary, take the union set to obtain: on x [-, the range of the value of x that matches the question is (-3 4, 4), so the range of the value of x in the original question is .

2kπ2kπ)kz

This is a proof of the inequality of the function class, and to treat this type of problem is to construct.

Take advantage of monotonicity.

Prove. You wrote the title incorrectly, the x on the left

It should be molecular.

Solution: On the left side of the precursor, let f(x) sinx x, to prove sinx x, as long as f(x) is proved to be 0, which is equivalent to the maximum value of f(x) on (0, 2) is less than 0

Derivation. f'(x) cosx 1 when 0 opens.

Therefore, g(x) g(0) 0, i.e., 2x sinx, we can see that 2x sinx x

Let f(x)=x-sinx

g(x)=sinx-2/πx

Then there is: f'(x)=1-cosx>=0

g'(x)=cosx-2/π

So f(x) is an increasing function.

f(x)>f(0)=0

i.e.: x>sinx

So g(x) increases first and then decreases, and the minimum value is g(0) or g(2) which is calculated to be 0 for both, so g(x) > 0

i.e. sinx>2 x

So: 2 x

The most interesting proof would be to use the definition in the unit circle, where sinx is the chord length and x is the arc length.

wqnjnsd

The proof is also very straightforward, but it can be a little stricter, and it is okay to be wordy

When x>1, it is clearly true.

When 00, i.e. f(x).

is a monotonic increment.

f(0+)0-sin0

So. f(x)>0, ie. sinxx

This question. First of all, the function takes the maximum value of x, which must be x=1 2, and there is no doubt about this milliorange hail

As for your second line of thought, I don't really understand the meaning of Yuanfan.

But I can write a solution based on your ideas.

ab<=(a+b)/2)^2

When this reed is closed, ab if and only if a=b, take the maximum value, ((a+b) 2) 2 So the formula you wrote, there are 3x(1-x)<=3*(1 2) 2<=3 4 When this maximum value is taken 3 4, x=1-x, that is, x=1 2 I don't know if you can understand this explanation?

If you have any questions, just pm me

Hello. You can see it by making an image.

When x=4, sinx=cosx

When 0 answers for you, please ask if you don't understand, please choose to be satisfied in time if you understand! (*Thank you!)

Authoritative answer:

丨x-1丨+丨2-x丨 7

Get: x-1=0, 2-x=0.

Find the range in three intervals: when 17

So it is not true, and it is not advisable to be in this interval.

x>2, 丨x-1丨+丨2-x丨 7

x-1-(2-x)>7

2x>10

x>5 is desirable in this range.

When x<1.

x-1)+2-x>7

2x+3>7

2x>4

x<-2

It is advisable in this range.

In summary, the value range of x is as follows:

x>5 and.

x<-2

Solution: Because of the sine function y

sinx and cosine function y

The minimum positive period of cosx is 2, so consider the case on x [-, plus 2k and k z is sufficient. A discussion of the classification of the value of x yields:

1) If cosx

0, i.e. x (-2, 2), so sinx

cosxtanx

x∈(-2kπ，π4

k ), k z, so when k

0, x2) if cosx

0, sinx at this time

1, by the title sinx

1, at this time x3) if cosx

0, i.e. x (-2) (2, ) so sinxcosx

tanxx∈(π4

k , 2k ), k z, to make x (-2) (2, ) so when k-1, x

To sum up, taking the union set obtains: on x [-, the range of values of x that fits the meaning of the question is (-3 4, 4), so the range of values of x sought in the original question is.

2kπ，π4

2kπ)，kz

ln x is monotonically increasing, ln1 = 0 lne=1

So 1<=x<=e

Draw an image of the sine function on [- 2, Qiaoqiao 3 2]. Because -1a

Therefore, the value range of x is .

arcsin(-a),arcsin(-a))

a<0 so [0, ] definitely does.

So Kuan Shi looks at the third.

The four elephants are limited to the virtual book.

a<0 so-2 so and him about y

Axial symmetry. of, in.

The third difference macro quadrant.

is 2K - Arcsina

In range is -arcsina

So it's [arcsina, -arcsina] to choose b