Trigonometric problems, which require a process

y= sin(π/6-2x)-cos2x

sin2xcosπ/6+cos2xsinπ/6)-sin(2x+π/6)

Minimum positive period: 2 2 =

When 2x+6=2k-2, the maximum ymax=1 at this time x = k-3, where x belongs to z

We haven't learned trigonometry yet, and we'll learn it in a week.

First of all, it should be satisfied that the formula under the root number should be greater than 0, that is, 1+2cos2a 0 1-2cos2a 0 to get a belongs to [ 6, 3] or [2 3, 5 6] or [7 6, 4 3] or [5 3, 11 6].

Secondly, the inequality is squared at the same time, so that 4sin a 2-2 (1-4cos a) 4, 4sin a 2-2 (1-4cos a) can be deduced to obtain (1-4cos a) cos 2a deduced (cos2a-1) 0 constantly.

2-2 (1-4cos a) 4 can be deduced (1-4cos a) -2 constantly.

In summary, the result should be that a belongs to [6, 3] or [2 3, 5 6] or [7 6, 4 3] or [5 3, 11 6].

The part of the absolute value is squared to get 2-2*under the root sign (1-4*cos2a 2), and then the term is shifted and simplified to prove the new inequality: 2*sina 2<=1-under the root sign (1-4*cos2a 2)<=2. To the left of the proband, 1-2*sina 2=cos2a.

After the two sides are shifted, 1-4*cos2a 2<=cos2a 2, so cos2a>=1 5The right one is obviously true, because [under the root number (1+2cos2a)]<2....Did you make a mistake???

It's all like this...

For af bc, then de is better than af=cd than ac=1 to 3, so. af=3de

and ab=4de

So sinb = 3de is better than 4de = 3 to 4

1. sin -cos = 1 2 square on both sides to get sin *cos = 3 8, sin +cos = 1 + 2 * 3 8 = 7 2

2. Because sin -cos = 1 2, sin +cos = 7 2, so sin = (1+ 7) 4, cos = (1 + 7) 4, tan = sin cos = (4 + 7) 3

sin4 to the power -cos4 +tan = (sin2 to the power + cos2 ) * sin2 to the power -cos2 ) + tan

sinθ+cosθ）*sinθ-cosθ）+tanθ=7/12*√7+4/3

1: The note f(x)=sin2x+acos2x is symmetrical from the image of y=sin2x+acos2x with respect to the straight line x=8:

f(-4)=f(0), i.e.: -1=a

2: To correct the problem, the definition domain of f(x) should be negative infinity to zero and zero to positive infinity, otherwise f(0)=0, and f(1 2)=0 from the properties of the odd function, so f(x) in (- not monotonic contradicts the title.

Solution: From the properties of the odd function: f(1 2) = f(-1 2) = 0, and f(x) also increases monotonically at (- 0).

Obtained from the images of f(cosa) 0 and f(x).

cosa -1 2 or 0 cosa 1 2 again because a is the inner angle of a triangle.

So a can be in the range of (0, 3) or (2 3), your answer is incorrect.

Regarding the origin symmetry, it is an odd function, and cosx is not an odd function, sinx is an odd function, and cos(x+ 2+k)=sinx, so we want to express the following cos in the form of 2+k, so the same as below.