Find the range of x that satisfies sin x 4 1 2 process how that intersection came out

Method 1: Use the unit circle to find it.

Draw a unit circle, and then draw a horizontal line y=1 2;The angles of the two intersections are 6 and 5 6;

Satisfying less than or equal to 1 2 is below the straight line, to write the form of the interval, we must follow the counterclockwise, so, we can write 6 as 13 6, then 5 6 + 2k x + 4 13 6 + 2k ;

Goat: 7 12+2k x 23 12+2k ;

You can also write 5 6 as -7 6, then -7 6+2k x+ 4 6+2k ;

-17 12+2k x - 12+2k ;

Both answers are fine.

Method 2: Use the trigonometric image, i.e., the sinusoidal curve.

Draw an image of a sinusoidal curve on [0,2 ], and draw a horizontal line y=1 2 with intersections of 6 and 5 6;

Satisfying less than or equal to 1 2 is below the straight line, with two intervals [0, 6] and [5 6,2];

So: 2k x+ 4 6+2k or 5 6+2k x+ 4 2 +2k ;

- 4+2k x - 12+2k or 7 12+2k x 7 4+2k ;

Both approaches are possible, and if you want to ask how 6 and 5 6 came about, I can only say that this is basic common sense, which needs to be mastered, and can be deduced in the unit circle.

Because sin ( 6) = 1 2

So x+ 4 = 6 +2k (k n).

Do the rest for yourself (draw the image, combine the math, and you can write the range).

y=sin(x) is the odd function Wang Rang.

So sin(-x) = -sin(x).

sin (-x)=(sinx) trap = sin debate x<>

Since sin(-x)=-sin(x), sin(-x)=(sin(x)) sin(x). The specific derivation process is as follows:

Let x be any angle, then there is sin(-x)=-sin(x).

The square of the left and right sides of the above equation is obtained by sin (-x)=(sin(x)) The lead excavation sedan is scattered in (-a)(-b)=ab, so there is (-sin(x)) sin (x).

i.e. sin (x) sin x ) is proven.

According to the periodic referentiality of the sinusoidal function, sin(x) = sin(x + 2), therefore:

sin²(-x) =sin²(-x + 2π) sin²(2π -x) =sin²(x - 2π) sin²x

Therefore, sin is only mu pei (-x) = sin x.

Generally speaking, for trigonometric functions, when we study the values and properties of formulas like sinx+cosx, because the change laws of the two formulas are not consistent, we should unify them into one formula.

2(√2/2sinx+√2/2cosx)=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)

The calculation process is as above. In fact, it is the inverse application of the sine formula of the sum and difference of the two angles of the trigonometric function. The basic types can be summarized as:

asinx+bcosx= a 2+b 2sin(x+y), where y can be solved by its satisfaction of tany=b a. Hope it helps.

Your results should be marked with a minus sign.

Constructing a right triangle ABC, right angle C, let the inner angle of angle A be denoted as , the inner angle of angle B denoted as , and the outer angle of angle b = 2+ then: sin

/2+α)sin

sinβ=cos

cos = -1 3, sin(x+ )7 9, and x (0, 2), 2, ) pants.

So sin = 2 2 3, cos(x+ )4 2 9

cosx=cos[(x+β)

cos(x+β)

cosβ+sin(x+β)sinβ

sinx 2+cosx 2=(2 root number 3) 3, the square of both sides gets 1+2sinx 2cosx Hu shouts 2=4 3,2sinx 2cosx 2=1 3,sinx=1 3

cos2x=1-2sin x=7 Nian Oak 9

Share a solution. x [-5,5], the integrand is sensitive to the simple x sin x x x 4.

The number of odd letters.

Root bridge pants according to. Definite integrals.

property, primitive=0.

FYI.

There is a problem with your solution: the original problem is to define the range of the value of x [ 8, 3 4], and you are now shilling 2x+3 4=2k + 2, so are you sure that the function y=sin(2x+3 4) in this problem must be able to take the maximum value at 2k + 2? You see x is in [ 8, 3 4], so 2x+3 4 is in [ ,9 4] and does not contain 2k + 2, so it is not the maximum value at 2k + 2.

The minimum value is OK.

Let 2x+3 4=2k + 2, solve the maximum, and get x=k - 8, on the interval [ 8, 3 4], so k is not.

On the interval [8,3 4], that is, 2x+3 4 is on [,9 4], find the maximum value in this interval.

sin2x=cos(π/2-2x)

cos[2(π/4-x)]

1-2sin²(π/4-x)

I hope my answer is helpful to you.