An angle x has sec x < 0 and cot x > 0. In which quadrant does x lie?

Step 1. Cosine. $\sec x < 0$ means $\cos x < 0$: quadrant II or III.

Step 2. Tangent. $\cot x > 0$ means $\tan x > 0$: quadrant I or III.

Step 3. Both. Only quadrant 3 (III) fits.

Step 4. The angle. The acute angle with sine $\dfrac{1}{2}$ is $\dfrac{\pi}{6}$. In quadrant III, $x = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}$.