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I have $\cos{Q}=-0.5$, required to find $180^{\circ}\leq Q\leq 360^{\circ}$.

What I tried: The acute angle is $60^{\circ}$, then, since cosine is negative on third quadrant, I suppose the value is $180^{\circ}+60^{\circ}=240^{\circ}$. Am I right?

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You are correct. There are two values within a circle that have a given cosine (except -1,0,1). For $\cos Q=0.5$, these are $150^\circ$ and $240^\circ$ and you have found the right one.