Find the maximum and minimum of y = 5cos & # 178; X + 3sin & # 178; x-2sinxcosx, when x ∈ [0, π / 2], the maximum and minimum of the function

Find the maximum and minimum of y = 5cos & # 178; X + 3sin & # 178; x-2sinxcosx, when x ∈ [0, π / 2], the maximum and minimum of the function

The solution y = 5cos & # 178; X + 3sin & # 178; x-2sinxcosx
=3cos²x+3sin²x+2cos²x-2sinxcosx
=3+2cos²x-2sinxcosx
=3+1+cos2x-sin2x
=cos2x-sin2x+4
=√2（√2/2cos2x-√2/2sin2x）+4
=√2cos（2x+π/4）+4
From X ∈ [0, π / 2]
That is, 0 ≤ x ≤ π / 2
That is, 0 ≤ 2x ≤ π
That is, π / 4 ≤ 2x + π / 4 ≤ 5 π / 4
That is, when 2x + π / 4 = π / 4, the maximum value of Y is √ 2 × √ 2 / 2 + 4 = 5
When 2x + π / 4 = π, the maximum value of Y is √ 2 × (- 1) + 4 = 4 - √ 2
Given f (x) = - 3sin & # 178; x-4cosx + 2. Find the maximum and minimum of F (x)
（2）2sin²a-sinacosa+cos²a
f（x）=-3sin²x-4cosx+2=3-3sinx^2-4cosx+2=3cosx^2-4cosx+2
The axis of symmetry of the function is cosx = 2 / 3
-1
How to solve the equation with the collocation method
2x2 + 3 = 7x2x2 (7) x + 3 = 0x2 (7) / 2x + 3 / 2 = 0x2 (7) / 2x + (7 / 4) 2 - (7 / 4) 2 + 3 = 0 (x (7) / 4) 2 (1) / 16 = 0 (x (7) / 4) 2 = 1 / 16x (7) / 4 = ± 1 / 4x = ± 1 / 4 + 7 / 4x = 2 or x = 3 / 2. If you have any questions, please use it! Check the original post > >
When x = ----, the minimum value of the algebraic formula x ^ 2-2x + 2 is
Can only be 0, minimum 2
x^2-2x+1+1
=(x-1)^2+1
When x = 1, the minimum value of the original formula is 1
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Original formula = (x-1) ^ 2 + 1
When x = 1 is, there is a minimum value of 1
Let x > 1 / 2, then the minimum value of the algebraic formula x + 8 / (2x-1) is
X>1/2
2x-1>0
X+8/(2X-1)
=（2x-1)/2+8/(2x-1)+1/2
≥2√[(2x-1)/2*8/(2x-1)] +1/2
=2*2+1/2
=9/2
The minimum value is 9 / 2