Recall that `sinx + cosx =u` (say)
and `sin x cosx =v` (say) are connected by
`(sinx +cosx)^(2) = sin^(2)x + cos^(2)x+2sin cosx`
`rArr u^(2) = 1+2v`
`rArr v=(u^(2)-1)/(2)`
It follows that any rational integral function of `sinx + cosx`, and `sinx cosx` i.e., `R(sinx + cosx, sinx cosx)`, or in our notation R(u,v) can be transformed to `R(u, (u^(2)-1)/2)`. Thus, to solve an equation of the form `R(u,v)=0`, we form a polynomial equation in u and than look for solutions.
The solution set of
`sinx + cosx -2sqrt(2) sin x cosx=0` is completely described by

Options

A. `x=2npi +pi/4, 2npi+ (11pi)/12, n int Z`
B. `x =2npi -pi/4, 2npi + (7pi)/12, n int Z`
C. `x=2npi + pi/4, 2npi -(7pi)/12, n int Z`
D. `x = 2npi- pi/4, 2npi-(7pi)/12, n int Z`

Answer

To solve the equation \( \sin x + \cos x - 2\sqrt{2} \sin x \cos x = 0 \), we will follow these steps:

### Step 1: Define Variables
Let:
- \( u = \sin x + \cos x \)
- \( v = \sin x \cos x \)

### Step 2: Rewrite the Equation
The given equation can be rewritten as:
\[ u - 2\sqrt{2}v = 0 \]
This implies:
\[ u = 2\sqrt{2}v \]

### Step 3: Square Both Sides
Now, square both sides:
\[ u^2 = (2\sqrt{2}v)^2 \]
This simplifies to:
\[ u^2 = 8v^2 \]

### Step 4: Use the Identity
We know from the identity:
\[ u^2 = 1 + 2v \]
Now we can set the two expressions for \( u^2 \) equal to each other:
\[ 1 + 2v = 8v^2 \]

### Step 5: Rearrange the Equation
Rearranging gives us a quadratic equation:
\[ 8v^2 - 2v - 1 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 8, b = -2, c = -1 \).
\[ v = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
\[ v = \frac{2 \pm \sqrt{4 + 32}}{16} \]
\[ v = \frac{2 \pm \sqrt{36}}{16} \]
\[ v = \frac{2 \pm 6}{16} \]

### Step 7: Calculate the Values of \( v \)
Calculating the two possible values:
1. \( v = \frac{8}{16} = \frac{1}{2} \)
2. \( v = \frac{-4}{16} = -\frac{1}{4} \)

### Step 8: Relate Back to \( \sin x \) and \( \cos x \)
Recall that \( v = \sin x \cos x \):
1. For \( v = \frac{1}{2} \):
\[ \sin x \cos x = \frac{1}{2} \implies \sin 2x = 1 \]
This gives:
\[ 2x = \frac{\pi}{2} + 2n\pi \implies x = \frac{\pi}{4} + n\pi \]

2. For \( v = -\frac{1}{4} \):
\[ \sin x \cos x = -\frac{1}{4} \implies \sin 2x = -\frac{1}{2} \]
This gives:
\[ 2x = -\frac{\pi}{6} + 2n\pi \implies x = -\frac{\pi}{12} + n\pi \]
and
\[ 2x = \frac{7\pi}{6} + 2n\pi \implies x = \frac{7\pi}{12} + n\pi \]

### Final Solution
The complete solution set for \( x \) is:
\[ x = \frac{\pi}{4} + n\pi, \quad x = -\frac{\pi}{12} + n\pi, \quad x = \frac{7\pi}{12} + n\pi \]