Find the solutions `x in [0,2pi]` of:`"sin" 2x – 12 ("sin" x - cos x) + 12 = 0`

Answer

To solve the equation \( \sin 2x - 12(\sin x - \cos x) + 12 = 0 \) for \( x \) in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation using the double angle identity for sine:
\[ \sin 2x = 2 \sin x \cos x \]
Substituting this into the equation gives:
\[ 2 \sin x \cos x - 12(\sin x - \cos x) + 12 = 0 \]

### Step 2: Expand and rearrange
Expanding the equation:
\[ 2 \sin x \cos x - 12 \sin x + 12 \cos x + 12 = 0 \]
Rearranging it, we have:
\[ 2 \sin x \cos x - 12 \sin x + 12 \cos x + 12 = 0 \]

### Step 3: Substitute \( t = \sin x - \cos x \)
Let \( t = \sin x - \cos x \). Then we can express \( \sin x \) and \( \cos x \) in terms of \( t \):
\[ \sin x = t + \cos x \]
Squaring both sides gives:
\[ \sin^2 x = (t + \cos x)^2 = t^2 + 2t \cos x + \cos^2 x \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ 1 = t^2 + 2t \cos x + \cos^2 x \]
This leads to:
\[ 1 - t^2 = 2t \cos x \]

### Step 4: Substitute back into the equation
Substituting \( \sin 2x = 2 \sin x \cos x \) into the equation gives:
\[ 1 - t^2 - 12t + 12 = 0 \]
This simplifies to:
\[ -t^2 - 12t + 13 = 0 \]
Multiplying through by -1:
\[ t^2 + 12t - 13 = 0 \]

### Step 5: Factor the quadratic
We can factor the quadratic:
\[ (t + 13)(t - 1) = 0 \]
Thus, the solutions for \( t \) are:
\[ t = -13 \quad \text{or} \quad t = 1 \]

### Step 6: Analyze the solutions
1. For \( t = -13 \):
\[ \sin x - \cos x = -13 \]
This is impossible since both \( \sin x \) and \( \cos x \) are bounded between -1 and 1.

2. For \( t = 1 \):
\[ \sin x - \cos x = 1 \]
Rearranging gives:
\[ \sin x = 1 + \cos x \]
Since \( \sin x \) cannot exceed 1, we analyze this case further.

### Step 7: Solve \( \sin x = 1 + \cos x \)
This leads to:
\[ \sin x - \cos x = 1 \]
Multiplying both sides by \( \sqrt{2} \):
\[ \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}} \]
This can be rewritten using the sine subtraction formula:
\[ \sin\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, we have:
\[ x - \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4} \]

### Step 8: Find specific solutions for \( x \)
1. For \( n = 0 \):
\[ x - \frac{\pi}{4} = \frac{\pi}{4} \implies x = \frac{\pi}{2} \]

2. For \( n = 1 \):
\[ x - \frac{\pi}{4} = \pi - \frac{\pi}{4} \implies x = \frac{5\pi}{4} \]

### Step 9: Final solutions
The solutions for \( x \) in the interval \( [0, 2\pi] \) are:
\[ x = \frac{\pi}{2}, \frac{5\pi}{4} \]