Find `max^(m)&min^(m)` value of `10cos^(2)x-6sinx.cosx+2sin^(2)x`

Answer

To find the maximum and minimum values of the expression \(10 \cos^2 x - 6 \sin x \cos x + 2 \sin^2 x\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 10 \cos^2 x - 6 \sin x \cos x + 2 \sin^2 x \]
We can rearrange it as follows:
\[ = 10 \cos^2 x + 2 \sin^2 x - 6 \sin x \cos x \]

### Step 2: Combine terms
Notice that \(10 \cos^2 x + 2 \sin^2 x\) can be expressed as:
\[ = 9 \cos^2 x + 2 (\sin^2 x + \cos^2 x) = 9 \cos^2 x + 2 \]
Using the identity \(\sin^2 x + \cos^2 x = 1\).

### Step 3: Factor the quadratic form
Now, we can express the quadratic in terms of \(\cos x\) and \(\sin x\):
\[ = 9 \cos^2 x - 6 \sin x \cos x + 2 \sin^2 x \]
This can be rewritten as:
\[ = (3 \cos x)^2 - 2(3 \cos x)(\sin x) + (\sin x)^2 + 1 \]
This resembles the form \(a^2 - 2ab + b^2\), which is \((a - b)^2\).

### Step 4: Complete the square
We can complete the square:
\[ = (3 \cos x - \sin x)^2 + 1 \]

### Step 5: Find the minimum value
The minimum value of a square term \((3 \cos x - \sin x)^2\) is \(0\). Therefore, the minimum value of the entire expression is:
\[ 0 + 1 = 1 \]

### Step 6: Find the maximum value
To find the maximum value, we need to determine the maximum of \((3 \cos x - \sin x)^2\). The maximum value of \(3 \cos x - \sin x\) can be found using the formula for the maximum of \(a \sin x + b \cos x\):
\[ \sqrt{a^2 + b^2} \]
Here, \(a = -1\) and \(b = 3\):
\[ \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
Thus, the maximum value of \((3 \cos x - \sin x)\) is \(\sqrt{10}\).

### Step 7: Calculate the maximum value of the expression
The maximum value of the expression is:
\[ (\sqrt{10})^2 + 1 = 10 + 1 = 11 \]

### Final Result
Thus, the maximum and minimum values of the expression \(10 \cos^2 x - 6 \sin x \cos x + 2 \sin^2 x\) are:
\[ \text{Maximum value} = 11, \quad \text{Minimum value} = 1 \]