\( \frac { 2 \sin x \cos x } { \sin x - \cos x } \geq 0 \)

To tackle the inequality \( \frac{2 \sin x \cos x}{\sin x - \cos x} \geq 0 \), let's first recognize that \( 2 \sin x \cos x \) is the same as \( \sin(2x) \). This means we can rewrite our inequality as \( \frac{\sin(2x)}{\sin x - \cos x} \geq 0 \). Next, we need to determine the intervals where both the numerator and denominator have the same sign (both positive or both negative) since a fraction is non-negative when both parts are positive or both are negative. Analyze the signs of \( \sin(2x) \) and \( \sin x - \cos x \) over their periodic cycles, which helps us find the solution sets. For the numerator \( \sin(2x) \), it's positive when \( 2x \) is in the intervals \( (0, \pi) + k\cdot 2\pi \) and negative when \( 2x \) is in \( (\pi, 2\pi) + k\cdot 2\pi \). For the denominator \( \sin x - \cos x \), it will change sign based on the values of \( \sin x \) and \( \cos x \). Setting this equal to zero yields solutions like \( x = \frac{\pi}{4} + k\cdot \frac{\pi}{2} \). Putting it all together involves testing critical points and determining which intervals satisfy the overall inequality. Happy graphing!