What is a monotonically decreasing interval of the function y sin2x?

Analytically, the function f(x) = root number x is incrementing from 0 to positive infinity.

Let g(x)=sinx be a sinusoidal function.

Let's talk about the original formula, and first find the definition domain, because the number under the root number cannot be negative.

That is, 3x is greater than or equal to 6 2k and less than or equal to 5 6 2k simplification to find the defined domain, and x must be taken within the defined domain.

Then we start talking about decreasing intervals.

Let 1 new function be f(x)=f[g(x)], if f(x) is an increasing function, then g(x) is a decreasing function.

So just find the decreasing function interval of 2sin(3x)1.

Then the value of the calculated value intersects with the defined domain.

I'm sorry I can only give you an idea, I'm too lazy, so I don't calculate. Thank you for your understanding.

kπ.Solution 2; 2cos2x+1/.

kπ－π/2kπ,k∈z;2. The decreasing interval of the function is the increasing interval of cos2x; x<,kπ),k∈z;2

y′=2sinxcosx=sin2x;2

0,k z, i.e. sin2x<:y=sin 2x= 1 2<, by 2k

The monotonically decreasing interval of the function y=sin 2x is (k :(k 2k ; 2x<.

π/4+kπ,3π/4+kπ)(k∈z)

For the function y=f(x)=sin2x, when the slag envy 2x ( 2+2k ,3 2+2k )(k z), f(x) decreases monotonically, and when x ( 4+k ,3 4+k )(k z) is solved, f(x) decreases monotonically, that is, the monotonically decreasing interval of the function y=sin2x is ( 4+k ,3 4+k )(k z).

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What is the monotonically decreasing interval of y=sin(-2x)?

y=sin(-2x)=-sin2x

Single increase range: balance and deficiency blind from 2+2k

Sinx Zonal or Mausoleum Belt is (2k - 2, 2k + 2).

The minus interval is (2k + 2, 2k + 3 2).

So here is 2k - 2

by introducing auxiliary angles; That is, the square root of the sum of squares of the two coefficients is extracted and the source is a trigonometric function of one angle;

y=2[(1/2)sin(1/2)x+(√3/2)cos(1/2)x]

2[sin(1/2)x*cos(π/3)+cos(1/2)x*sin(π/3)]

2sin[(1/2)x+(π3)]

Then put the horns; (1 2)x+( 3) substitution into the monotonic reduction interval of the standard sine to remove the leakage brother state and solve x is as follows:

2)+2kπ≤(1/2)x+(π3)≤(3π/2)+2kπ

6)+2kπ≤(1/2)x≤(7π/6)+2kπ

3)+4kπ≤x≤(7π/3)+4kπ

So the monotonic reduction interval of the original function is:

3) +4k ,7 3) +4k ] x z), 2, the original function can:

y=2[1/2*sin(1/2x)+√3/2*cos(1/2x)]

2 [cos(3)sin(1 2x)+sin( dust regret 3)cos(1 2x)].

2sin(π/3+1/2x)

Then the discussion will do. ,0,y=2(1/2sin1/2x++√3/2cos1/2x)=2sin(1/2x+π/3），2+2kπ≤ 1/2x+π/3≤3π/2+2kπ

The solution is 3+4k x 7 3 +4k k belongs to z,0,

Because x [0, ] carries a pin of 2x [0,2], when this envy 2x [ 2,3 2], i.e. x [ 4,3 4], the function y=sin2x is a subtraction function, i.e.

The monotonic decreasing interval of the function y=sin2x,x [0, ] is [ 4,3 4].

The period of the function is 2

As long as the crack expands from 0,2.

When the sensitivity is x [0, ], y=sinx increases monotonically, and the left half monotonically increases and the right half decreases monotonically.

When x ( 2 ], y=0 does not monotonically defeat the bridge source.

The above conclusion is actually the second quadrant monotonic reduction, that is, the monotonic reduction interval is:

2+2kπ,π2kπ】

y=sin(-2x+π/3)

sin(2x-π/3)

Then 2k - 2 2x- 3 2k + 22k - calendar 2+ 3 2x 2k + 2+ 32k - 6 2x 2k +5 6

kπ-π12≤x≤kπ+5π/12

Monotonically decreasing interval limb carrying combustion is [k - 12,k +5 12] (k z),1,

That is, find the monotonically increasing interval of the function y=sinx.

The trigonometric function is used to reduce the monotonic recursive detribution interval of the original function of the clear image (or trigonometric line) of the positive split number as.

2kπ-π2,2kπ+π2] ,k∈z