The minimum positive period of the function y = cos (2x tt / 3) + 1 is

The minimum positive period of the function y = cos (2x tt / 3) + 1 is

T=2pai/2=pai

If the function f (x) = sin (x + a) △ 3 A is between 0 and 2 in the closed interval, it is even function. Find a
If the function f (x) = sin (x + a) △ 3 A is between the closed interval 0 and 2, it is an even function. The problem of finding a is why, because it is an even function, we can get f (0) = positive and negative 1

If the function f (x) = sin (x + a) △ 3 A is between the closed interval 0 and 2, it is even function. The problem of finding a is why, because it is even function, we can get f (0) = plus minus 1. We know that sine function is odd function, cosine function is even function, f (x) = sin (x + a) / 3, on the surface, f (x) is sine function

Find the maximum and minimum values of the following functions and write the set of X when the maximum and minimum values are obtained: y = √ 2 + SiNx / open, (x ∈ R) y = 3-2cosx (x ∈ R)

y＝√2﹢sinx/π
When y max = √ 2 + 1, SiNx / π = 1, X / π = 2K π + π / 2, x = 2K π & # 178; + π & # 178 / 2, so x ∈ {Xi x = 2K π & # 178; + π & # 178 / 2, K ∈ Z}
When y is minimum = √ 2-1, SiNx / π = - 1, X / π = 2K π - π / 2, x = 2K π & # 178; - π & # 178 / 2, so x ∈ {Xi x = 2K π & # 178; - π & # 178 / 2, K ∈ Z}
y＝3-2cosx
When y max = 5, cosx = - 1, x = 2K π + π, so x ∈ {Xi x = 2K π + π, K ∈ Z}
When y min = 1, cosx = 1, x = 2K π, so x ∈ {Xi, x = 2K π, K ∈ Z}
(⊙ o ⊙ )

The set of X for finding the maximum and minimum value of the function y = - cos ^ 2x SiNx + 1 and finding the maximum and minimum value of the function y = - cos ^ 2x SiNx + 1

y=—cos^2x-sinx+1=sin^2x-sinx=(sinx-1/2)^2-1/4
max:2 sinx=-1 x=2kπ-π/2
min:-1/4 SiNx = 1 / 2 x = 2K π + π / 6 or 2K π + 5 π / 6