The maximum value of function y = sin ^ 2x cosx

The maximum value of function y = sin ^ 2x cosx

Find the value of a when the maximum value of F (x) = sin ^ 2x + α cosx-1 / 2 α - 3 / 2 (x ∈ R) is 1

f(x)=sin^2x+αcosx-1/2α-3/2
=1-cos^2 x+acosx-a/2-3/2
=-cos^2 x+acosx-a/2-1/2
=-[cosx-a/2)^2+(a^2)/4-a/2-1/2

Finding the maximum value of the function y = cosx sin ^ 2x + 7 / 4
And 2cos10 · - sin20 · / cos20 ·

Simplify y = cosx - [1 - (cosx) ^ 2] + 7 / 4
=(cosX)^2+cosX+3/4
Let t = cosx so - 1=

∫ - D (cosx) / 2-cos ^ 2x = - 1 / 2 radical sign 2 * ln (radical 2 + cosx) / (radical 2-cosx), how to calculate?
One more,
∫ (1 + cos2t) DT = a ^ 2 / 2 (T + 1 / 2sin2t) + C

The first question is a bit confusing, and the second one is ∫ (1 + cos2t) DT = ∫ (1) DT + ∫ (cos2t) DT = t + 1 / 2 ∫ (cos2t) D2T = t + sin2t + C

Given the vector a = (cos3 / 2x, SIN3 / 2x), B = (- SiNx / 2, - cosx / 2), where x belongs to [π / 2, π], 1. | a + B | = radical 3, find the value of X
2. Function f (x = a * B + | a + B | ^ 2, if C ＞ f (x) is constant, find the value range of real number C
The above two questions,

1. | a + B | ^ 2 = (cos3 / 2x - SiNx / 2) ^ 2 + (SIN3 / 2x - cosx / 2) ^ 2 = 3 (cos3 / 2x) ^ 2 - 2 * cos3 / 2x * SiNx / 2 + (SiNx / 2) ^ 2 + (SIN3 / 2x) ^ 2 - 2 * SIN3 / 2xcosx / 2 + (cosx / 2) ^ 2 = 3. By using sum of sine and cosine squares = 1 and the formula of sum and difference of products, we can get 2 - 2sin2

If 2Sin ^ 2x + sin ^ 2Y = 3sinx, then the value range of sin ^ 2x + sin ^ 2Y is

2sin^2x+sin^2y=3sinx
Sin ^ 2Y = - 2Sin ^ 2x + 3sinx into sin ^ 2x + sin ^ 2Y
∴
sin^2x+sin^2y
=sin^2x-2sin^2x+3sinx
=-sin^2x+3sinx
=-(sin^2x-3sinx+9/4)+9/4
=-(sinx-3/2)^2+9/4
Maximum = - (1-3 / 2) ^ 2 + 9 / 4 = 2
Minimum = - (- 1-3 / 2) ^ 2 + 9 / 4 = - 4
The value range is [- 4,2]
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Given 3sin ^ 2x + 2Sin ^ 2Y = 2sinx, the value range of sin ^ 2x + Xin ^ 2Y is

3sin ^ 2x + 2Sin ^ 2Y = 2sinx ｛ 2Sin & ｛ 178; y = 2sinx-3sin & ｛ 178; X ≥ 0 ｛ 0 ≤ SiNx ≤ 2 / 3 ｛ 2 * (sin ^ 2x + Xin ^ 2Y) = 2Sin & ｛ 178; X + 2sinx-3sin & ｛ 178; X + 2sinx = - (sinx-1) & ｛ 178; + 1 ｛ SiNx = 0, 2 * (sin ^ 2x + Xin ^ 2Y) has a minimum value of 0

Cot ^ 2x = 2tan ^ 2Y + 1 find sin ^ 2Y + 2Sin ^ 2x
Please write down the process if you can

cot^2x=2tan^2y+1 ,
cot^2x+1=2tan^2y+2,
1/(sinx)^2=2/(cosy)^2,
(cosy)^2=2(sinx)^2,
1-(siny)^2=2(sinx)^2,
(siny)^2+2(sinx)^2=1.

When x belongs to [0, π / 3], f (x) = the maximum value of COS ^ 2x / (2sinxcosx + cos ^ 2x sin ^ 2x)

f/(1-2f)=-(cosx/(sinx-cosx))^2=-(1/(1-tanx))^2
If f is the largest, then f / (1-2f) is the largest and | 1 / (1-tanx) | is the smallest
The maximum value of F is 1

What is the period of x times the absolute value of cosx as a whole

There is no cycle