Given f (x) = 2sinx / 2cosx / 2 + 2 (cosx / 2) * 2-1, find the set of all real numbers x that make f (x) + F (x) '= 0 I have worked out the original formula F (x) = 2sinx + cosx f(x)'=2cosx-sinx Finally, SiNx cosx = 0 What is the set of X? And what did I do right? The result is that x is π / 4 + 2 π n, and N is a positive integer,

Given f (x) = 2sinx / 2cosx / 2 + 2 (cosx / 2) * 2-1, find the set of all real numbers x that make f (x) + F (x) '= 0 I have worked out the original formula F (x) = 2sinx + cosx f(x)'=2cosx-sinx Finally, SiNx cosx = 0 What is the set of X? And what did I do right? The result is that x is π / 4 + 2 π n, and N is a positive integer,

No, there is something wrong with the original formula F (x) = 2sinx + cosx. It should be f (x) = SiNx + cosx
Then f (x) '= cosx SiNx, f (x) + F (x)' = 2cosx = 0
X is π / 2 + K π and K is an integer

Given the function f (x) = 32sinx-12cosx, (x ∈ R), find the maximum value of F (x), and find the set of X & nbsp; when f (x) reaches the maximum value

When π (x + x) is the largest value of π (x + sin) = 1, we take π (x + sin) = 6

The known function f (x) = 2sinxcosx-2cosx ^ 2 + 1
1. Finding the maximum value of F (x) and the corresponding x value
2. If f (θ) = 3 / 5, find the value of Cos2 (π / 4-2 θ)

1、
f(x)=sin2x-cos2x
=√2(√2/2*sin2x-√2/2cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
So 2x + π / 4 = 2K π + π / 2, sin max = 1
2X + π / 4 = 2K π - π / 2, Sinmin = - 1
therefore
X = k π + π / 8, maximum = √ 2
X = k π - 3 π / 8, minimum = - √ 2
2、
f(θ)=sin2θ-cos2θ=3/5
square
sin²2θ+cos²2θ-2sin2θcos2θ=9/25
1-sin4θ=9/25
sin4θ=16/25
The original formula = cos (π / 2-4 θ)
=sin4θ
=16/25