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y=2cos²x-1+1=cos2x-1
The monotone increasing interval of cosx is (2k π - π, 2K π)
So 2K π - π

Find the maximum and minimum value of the following function y = cos squared x + 3sinx-2

Y = cos? X + 3sinx-2 = 1-sin? X + 3sinx-2 = - sin? X + 3sinx-1 = - (sinx-3 / 2) mm2 + 5 / 4 because - 1 ≤ SiNx ≤ 1, the maximum value is - (1-3 / 2) mm2 + 5 / 4 = 1, and the minimum value is - (- 1-3 / 2) mm2 + 5 / 4 = - 5. If you don't understand, have a good study

The solution of the equation sin2x = cosx in the interval (0,2 π) is? And the minimum value of the function y = 4-3sinx-cos ^ 2x is

Sin2x = cosx → 2sinxcosx = cosx → SiNx = 1 / 2 → x = π / 6 or 5 π / 6
The original formula = 4-3sinx cos ^ 2x = sin ^ 2x-sinx + 3 = (sinx-3 / 2) ^ 2 + 3 / 4
So when SiNx = 1, take the minimum value = 1

Given that the function f (x) = follows 3sinx cosx, X belongs to R ① find the minimum positive period and maximum value of FX, ② monotonically increasing interval of FX, ③ the minimum value of FX on [0, PI]

(1)
f(x)
= √3sinx - cosx
= 2((√3/2)sinx - (1/2)cosx)
= 2 sin(x-π/6)
Minimum positive period = 2 π
Max = 2
(2)
Monotone increasing interval
2nπ-π/2

The minimum positive period of the function f (x) = (3sinx-4cosx) cosx

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F (x) = √ 3 × (1 - cos (2x)) / 2 - (sin (2x)) / 2 (using the double angle formula) = √ 3 / 2 - (√ 3 / 2) cos (2x) - (1 / 2) sin (2x) = √ 3 / 2 - sin (π / 3) cos (2x) - cos (π / 3) sin (2x) = √ 3 / 2 - sin (π / 3 + 2x) (...)

The maximum value of the function y = √ 3sinx + cosx, X ∈ [- π / 2, π / 2] is y=√3sinx+cosx =2Sin (x + a), (where Tana = 1 / √ 3, i.e. a = π / 6) =2sin(x+π/6) Because x ∈ [- π / 2, π / 2] So (x + π / 6) ∈ [- π / 3, (2 π) / 3] So y (max) = 2 But how can I finally figure out that one time is equal to one and another is equal to √ 3?

(x+π/6)∈[-π/3,(2π)/3]
SiNx increases at (- π / 3, π / 2)
(π / 2,2 π / 3) decline
So the maximum is sin π / 2 = 1
So the maximum is 2 × 1 = 2

The minimum positive period of the function y = (SiNx cosx) cosx is

y=sinxcosx-cos²x
=1/2*sin2x-(1+cos2x)/2
=1/2*(sin2x-cos2x)-1/2
=√2/2*sin(2x-π/4)-1/2
So t = 2 π / 2 = π

Given that the vector a = {sin (x / 2 + π / 12),) cosx / 2}, B = {cos (x / 2 + π / 12), - cosx / 2}. X belongs to [π / 2, π], the function f (x) is an image According to the vector C = (m, n), so that the translated image is symmetrical about the origin, the vector C

f（x）=1/2*sin(x+π/6)-cos²(x/2)
=.=1/2 sin(x-π/6)-1/2
To make the translated image symmetrical about the origin
Then M = - π / 6 + K π,
n=1/2
That is, the vector C is (- π / 6 + K π, 1 / 2) k is an integer

It is known that the function FX is equal to sin (x minus 6) plus cosx x x, including R to find the value of F0

F (0) = sin (0-π / 6) + cos0 = sin (- π / 6) + cos0 = sin (- π / 6) + cos0 = - 1 / 2 + 1 + 1 = 1 / 2, if you want to ask the result of simplification, then: F (x) = sin (x - π / 6) + cosx x = sinxcos (π / 6) - cosxsin (π / 6) + cosx = sinxcos (π / 6) - cosxsin (π / 6) + cosx = sinxcos (π / 6) - (1 / 2) cosx = sinxcos (π / 6) + (1 / 2) cosx = sinxcos (π / 6) + (1 / 2) cosx = sinxcos (π / 6) cosx = SiNx = sinxcos (π / 6) + / 6