How to find the period of F (x) = 2Sin ^ 2-2 radical 3 * SiNx * cosx + 1 As in the title,

How to find the period of F (x) = 2Sin ^ 2-2 radical 3 * SiNx * cosx + 1 As in the title,

f(x)=2sin^2-2√3*sinX*cosX+1 π
=1-cos2X-√3*sin2X+1
=2-2sin(2X+ π/3)
T=2 π/2= π

Known 0 Math homework help users 2016-11-29 report Use this app to check the operation efficiently and accurately!

F (x) = (1 + 2 cosx ^ 2 - 1 + 8sinx ^ 2) / 2sinxcosx = (cosx ^ 2 + 4sinx ^ 2) / sinxcosx simultaneously divided by cosx ^ 2 = (1 + 4tanx ^ 2) / TaNx = 1 / TaNx + 4tanx ∵ 0 ﹤ x ﹤ π

How to simplify the function f (x) = sin2x cos2x to f (x) = √ 2Sin (2x - π / 4)

f(x)=sin2x-cos2x
=√2*[sin2x*(√2)/2 - cos2x*(√2)/2]
=√2*[sin2x*cos(π/4) -cos2x*sin(π/4)]
=√2*sin(2x-π/4)

The image symmetry axes of the functions f (x) = 3sin (Wx - π / 6) (W big 0) and G (x) = 2cos (2x + @) (0 small @ small pie) are exactly the same, and the value of G (Pie / 3) is the same

It's a state of the country (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π) is (π) and (π = π (2) and (π) is (π) and (π = π) is (π) and (π) is (π) is (π) and (π = 3) is (s) (x) = 2CO

The known function f (x) = 3sin (ω X - π) 6) If the symmetry axes of (ω > 0) and G (x) = 2cos (2x + φ) (0 < φ < π) are exactly the same, then G (π) 3) The value of is______ .

The symmetric axis equation of the function f (x) = 3sin (ω X - π 6) (ω > 0) is ω X - π 6 = k π + π 2, that is, x = k π ω + 2 π 3 ω, K ∈ Z

Vector a = (SiNx, 1) B = (cosx, - 1 / 2), find the value range of F (x) = (a + b) * B on {- Pai / 2,0} Be specific... Be specific... Let's have a process first!

(x) = (a + b) · B = A.B + (b) B = A.B + (b) | 2 = sinxcos X-1 / 2 + cosx ^ 2 + 1 / 4 = sin (2x) / 2 + (1 + cos (2x)) / 2-1 / 4 = sin (2x) / 2 + cos (2x) / 2 + 1 / 4 = (√ 2 / 2) sin (2x + π / 4) + 1 / 4x ∈ (- π / 2,0) namely: 2x + π / 4 ∈ (- 3 π / 4, π / 4) namely: namely: 2x + π / 4 ∈ (- 3 π / 4, π / 4) that is: namely: 2x + π / 4, π / 4, π / 4), namely: namely: namely: the sin (2x + π / 4) ∈ [- 1, √ 2 / 2)

Given the vector a = (SiNx, 2 √ 3sinx), B = (mcosx, - SiNx), define f (x) = a * B + √ 3, and x = π / 6 is the zero point of the function y = f (x) (1) Find the single point interval of function y = f (x) on R 2. If the function y = f (x + θ) (0

The vector vector a = (SiNx, 2 √ 3sinx), B = (mcosx, - SiNx), f (x) = a, B = (mcosx, - SiNx), f (x) = a ● B + √ 3 = msinxcosx X-2 √ 3sin 2x + 3 = 3 = m / 2 * sin2x + √ 3 (1-2sin 2, x) = m / 2sin2x + √ 3cos2x ∵ x = x = π / 6 is the zero point of function y = f (x), f (π / 6) = m / 2 * sin π / 3 + √ 3cos π / 3 = 3 = m / 3 = m / 2 * sin π / 3 + √ 3cos π / 3 = M = M = m 9 3 = √ 3 / 4 + √ 3 /

① F (x) = 2Sin ^ 2x + sinx-2; f (x) = SiNx / (2 + SiNx)

① Because f (x) = 2Sin ^ 2x + sinx-2 = 2 (SiNx + 1 / 4) ^ 2-17 / 8, and because 0 < SiNx < 1 and when SiNx = - 1 / 4, the value range is - 2 ≤ f (x) ≤ 33 / 16; (2) because f (x) = SiNx / (2 + SiNx) = ((SiNx + 2) - 2) / (2 + SiNx) = 1-2 / 2 + SiNx, and because 0 < SiNx < 1

Let x (x) = (2) x (x) = (2) = (2) = (2) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = Let f (x) = 2 * cos ^ 2 x + 2 * radical 3 * sinxcosx (x ∈ R), and find the minimum positive period of F (x)

A:
f(x)=2(cosx)^2+2√3sinxcosx
=cos2x+1+√3sin2x
=2*[(1/2)cos2x+(√3/2)sin2x]+1
=2sin(2x+π/6)+1
So:
Minimum positive period T = 2 π / 2 = π

The function f (x) = (radical 3) sinxcosx + (COS ^ 2) x + 1 is known Find its minimum positive period and monotone decreasing interval; in the triangle ABC, a, B, C are opposite sides of the angle respectively. If f (a) = 2, B = 1 and the area of triangle ABC is (radical 3) / 2, find the value of A

function
f(x)=(√3)sinxcosx+cos²x+1
=(√3/2)(2sinxcosx)+(1/2)(2cos²x)+1
=(√3/2)sin2x+(1/2)(1+cos2x)+1
=sin(2x)cos(π/6)+cos(2x)sin(π/6)+(3/2)
=sin[2x+(π/6)]+(3/2)
[[[1]]]
Easy to know,
T=π
[[[2]]]
2kπ+(π/2)≤2x+(π/6)≤2kπ+(3π/2)
The monotonic decreasing interval is
kπ+(π/6)≤x≤kπ+(2π/3)
Namely [K π + (π / 6), K π + (2 π / 3)]
[[[3]]]
f(A)=2
sin[2A+30º]+(3/2)=2
sin(2A+30º)=1/2
It is easy to know that 2A + 30 ° is 150 ° at this time
∴A=60º
Triangle area
S=(bc/2)sinA
(c/2)×(√3)/2=√3/2
∴c=2
From the cosine theorem we can get the conclusion
a²=b²+c²-2bccosA
=1+4-2=3
∴a=√3