If `x+y=3-cos 4 theta` and `x-y=4sin 2 theta`,the value of `sqrtx+sqrty` is equal to

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we start with the given equations:

1. \( x + y = 3 - \cos(4\theta) \) (Equation 1)
2. \( x - y = 4 \sin(2\theta) \) (Equation 2)

### Step 1: Add the two equations
We add Equation 1 and Equation 2:

\[ (x + y) + (x - y) = (3 - \cos(4\theta)) + (4 \sin(2\theta) \]

This simplifies to:

\[ 2x = 3 - \cos(4\theta) + 4 \sin(2\theta) \]

### Step 2: Solve for \( x \)
Now, divide both sides by 2:

\[ x = \frac{1}{2} (3 - \cos(4\theta) + 4 \sin(2\theta)) \]

### Step 3: Subtract the two equations
Next, we subtract Equation 2 from Equation 1:

\[ (x + y) - (x - y) = (3 - \cos(4\theta)) - (4 \sin(2\theta)) \]

This simplifies to:

\[ 2y = 3 - \cos(4\theta) - 4 \sin(2\theta) \]

### Step 4: Solve for \( y \)
Now, divide both sides by 2:

\[ y = \frac{1}{2} (3 - \cos(4\theta) - 4 \sin(2\theta)) \]

### Step 5: Find \( \sqrt{x} + \sqrt{y} \)
Now we need to find \( \sqrt{x} + \sqrt{y} \).

From our expressions for \( x \) and \( y \):

\[ \sqrt{x} = \sqrt{\frac{1}{2} (3 - \cos(4\theta) + 4 \sin(2\theta))} \]
\[ \sqrt{y} = \sqrt{\frac{1}{2} (3 - \cos(4\theta) - 4 \sin(2\theta))} \]

### Step 6: Combine the square roots
We can factor out \( \sqrt{\frac{1}{2}} \):

\[ \sqrt{x} + \sqrt{y} = \sqrt{\frac{1}{2}} \left( \sqrt{(3 - \cos(4\theta) + 4 \sin(2\theta))} + \sqrt{(3 - \cos(4\theta) - 4 \sin(2\theta))} \right) \]

### Step 7: Simplify the expression
Let \( A = 3 - \cos(4\theta) \). Then we have:

\[ \sqrt{x} + \sqrt{y} = \sqrt{\frac{1}{2}} \left( \sqrt{A + 4 \sin(2\theta)} + \sqrt{A - 4 \sin(2\theta)} \right) \]

Using the identity \( \sqrt{a + b} + \sqrt{a - b} = 2\sqrt{a} \) when \( a \geq b \):

\[ \sqrt{x} + \sqrt{y} = \sqrt{\frac{1}{2}} \cdot 2\sqrt{A} = \sqrt{2A} \]

Substituting back for \( A \):

\[ \sqrt{x} + \sqrt{y} = \sqrt{2(3 - \cos(4\theta))} \]

### Step 8: Final Calculation
Since \( \cos(4\theta) \) can vary, we can conclude that the maximum value of \( \sqrt{x} + \sqrt{y} \) occurs when \( \cos(4\theta) \) is minimized, which leads to:

\[ \sqrt{x} + \sqrt{y} = 2 \]

Thus, the final answer is:

\[ \sqrt{x} + \sqrt{y} = 2 \]