To solve the equation \( \sin x \cos x = -\frac{\sqrt{2}}{4} \) on the interval \([0, 2\pi)\), we can start by using a trigonometric identity. Notice that:
[tex]\[ \sin x \cos x = \frac{1}{2} \sin(2x) \][/tex]
Therefore, the equation transforms to:
[tex]\[ \frac{1}{2} \sin(2x) = -\frac{\sqrt{2}}{4} \][/tex]
We can simplify this further by multiplying both sides of the equation by 2:
[tex]\[ \sin(2x) = -\frac{\sqrt{2}}{2} \][/tex]
Next, we need to determine the values of \(2x\) that satisfy this equation. We know that \(\sin \theta = -\frac{\sqrt{2}}{2}\) at the angles:
[tex]\[ \theta = \frac{5\pi}{4} + 2k\pi \quad \text{and} \quad \theta = \frac{7\pi}{4} + 2k\pi \][/tex]
for any integer \(k\). Hence for \(2x\),
[tex]\[ 2x = \frac{5\pi}{4} + 2k\pi \quad \text{and} \quad 2x = \frac{7\pi}{4} + 2k\pi \][/tex]
Now, we divide by 2 to solve for \(x\):
[tex]\[ x = \frac{5\pi}{8} + k\pi \quad \text{and} \quad x = \frac{7\pi}{8} + k\pi \][/tex]
Next, we need to consider values of \(x\) that lie within the interval \([0, 2\pi)\).
For \( x = \frac{5\pi}{8} + k\pi \):
- When \( k = 0 \):
[tex]\[ x = \frac{5\pi}{8} \][/tex]
- When \( k = 1 \):
[tex]\[ x = \frac{5\pi}{8} + \pi = \frac{5\pi}{8} + \frac{8\pi}{8} = \frac{13\pi}{8} \][/tex]
- When \( k = 2 \):
[tex]\[ x = \frac{13\pi}{8} + \pi = \frac{13\pi}{8} + \frac{8\pi}{8} = \frac{21\pi}{8} \][/tex]
\( \frac{21\pi}{8} \) is not within the interval \([0, 2\pi)\) since \(2\pi = \frac{16\pi}{8}\).
For \( x = \frac{7\pi}{8} + k\pi \):
- When \( k = 0 \):
[tex]\[ x = \frac{7\pi}{8} \][/tex]
- When \( k = 1 \):
[tex]\[ x = \frac{7\pi}{8} + \pi = \frac{7\pi}{8} + \frac{8\pi}{8} = \frac{15\pi}{8} \][/tex]
- When \( k = 2 \):
[tex]\[ x = \frac{15\pi}{8} + \pi = \frac{15\pi}{8} + \frac{8\pi}{8} = \frac{23\pi}{8} \][/tex]
\( \frac{23\pi}{8} \) is also not within the interval \([0, 2\pi)\).
Thus, the solutions within the interval \([0, 2\pi)\) are:
[tex]\[ x = \frac{5\pi}{8}, \frac{13\pi}{8}, \frac{7\pi}{8}, \frac{15\pi}{8} \][/tex]
