Minimum positive period of function FX = sin (cosx)?

Minimum positive period of function FX = sin (cosx)?

Function f (x) = sin (cosx)
f(x+2π)
=sin[cos(x+2π)]
=sin(cosx)
=f(x).
ν 2 π is a period of the function
Suppose that f (x + T) = f (x), (x ∈ R, t is a positive constant)
﹣ there is always sin [cos (x + T)] - sin (cosx) = 0
Left sum difference product,
left
=2cos{[cos(x+T)+cosx]/2}sin{[cos(x+T)-cosx]/2}=0
=-2cos{cos(x+(T/2))cos(T/2)}sin{sin(x+(T/2))sin(T/2)}=0
It is easy to know that sin (T / 2) = 0 is necessary for the above formula to be 0
ν Tmin = 2 π
The minimum positive period of the function is 2 π

Find the value range of the function y = 2sinx + radical (1-2sinx)

The range of SiNx is obtained from the requirement of definition domain. If t = SiNx, the problem can be transformed into the problem of finding the range of quadratic function. Note that t = SiNx has requirements

0

1.y=-3cos2x-1
-pai/2

The range of y = (3cosx + 1) / (cosx + 2) x ∈ [- π / 2,2 π / 3] is

y=（3cosx+1）/（cosx+2）
ycosx+2y=3cosx+1
(y-3)cosx=1-2y
cosx=(1-2y)/(y-3)
x∈[-π/2,2π/3]
-1/2≤cosx≤1
-1/2≤(1-2y)/(y-3)≤1
(a)-1/2≤(1-2y)/(y-3)
(2-4y+y-3)/(2y-6)≥0 (3y+1)/(y-3)≤0
-1 / 3 ≤ Y3 or Y ≤ 4 / 3 (2)
To sum up (1) (2) - 1 / 3 ≤ y ≤ 4 / 3
That is, the range is [- 1 / 3,4 / 3]

If the equation SiNx + root 3cosx + 2a-1 = 0 has two unequal real roots on [0, π], find the value range of real number a

If the equation SiNx + (√ 3) cosx + 2a-1 = 0 has two unequal real roots on [0, π], find the value range of real number a SiNx + (√ 3) cosx = 1-2a2 [(1 / 2) SiNx + (√ 3 / 2) cosx] = 2 [sinxcos (π / 3) + cosxsin (π / 3)] = 2Sin (x + π / 3) = 1-2a, sin (x + π / 3) = (1-2a) / 2, because in [0

The value range of 0 ≤ x ≤ 2 π, SiNx > root 3cosx, X

1. cosx>0
TaNx > radical 3
π/32. cosx<0
TaNx < radical 3
π/2<=x<4π/3
therefore
x∈（π/3,4π/3）

If SiNx radical 3cosx = 2m-4 is meaningful, then the range of M

Let f (x) = SiNx radical 3cosx = 2 (SiNx / 2-radical 3cosx / 2) = 2Sin (x-pi / 3)
The value range of F (x) is [- 2,2]
So - 2=

If x belongs to R, let 3cosx 6-4m under SiNx radical, find the value range of M Is equal to 6-4m

My idea is separation of variables
It is written as M = (3cosx - SiNx) / 4
Then the extremum of the function on the right is calculated and the value range of M is deduced
Triangle master general, not very good at seeking >

If x belongs to R, let 3cosx = (4m-6) / 4-m under SiNx radical sign, and find the value range of M

pi=3.14
sinx-(3)^0.5*cos(x)=2sin(x-pi/3)=(4m-6)/(4-m) =>
-2

Let the equation SiNx + √ 3cosx = a have two distinct real roots X1 and X2 in the interval (0,2 π). Find the value range of a and the value of X1 + x2 This is the answer sinx+√3cosx=a sinx*1/2+√3cosx/2=a/2 sin(x+π/3)=a/2 When-2

sinx+√3cosx=a
sinx*1/2+√3cosx/2=a/2
sin(x+π/3)=a/2
When-2