When is $\sin x$ or $\sin 2x$ larger than a half?

When is $\sin x$ or $\sin 2x$ larger than a half?

Question

The fraction of the interval $0\leq x \leq 2\pi$, for which one (or both) of the inequalities $\sin x \geq\frac{1}{2}, \qquad \sin 2x \geq \frac{1}{2}$ is true, equals

1. $\dfrac{1}{3}$;

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1. $\dfrac{13}{24}$;

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1. $\dfrac{7}{12}$;

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1. $\dfrac{5}{8}$.

Solution

The fraction of the interval $0\leq x \leq 2\pi$, for which one (or both) of the inequalities $\sin x \geq\frac{1}{2}, \qquad \sin 2x \geq \frac{1}{2}$ is true, equals

1. $\frac{1}{3}$;

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1. $\frac{13}{24}$;

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1. $\frac{7}{12}$;

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1. $\frac{5}{8}$.

In the range $0\leq x\leq2\pi$ we have $\sin x\geq\dfrac{1}{2}$ if and only if $\dfrac{\pi}{6}\leq x \leq \dfrac{5\pi}{6},$

and $\sin 2x\geq\dfrac{1}{2}$ if and only if $\dfrac{\pi}{12}\leq x \leq \dfrac{5\pi}{12}\quad \text{or} \quad \dfrac{13\pi}{12} \leq x \leq \dfrac{17\pi}{12}.$

So at least one of the inequalities holds for $\dfrac{\pi}{12}\leq x \leq \dfrac{5\pi}{6}$ and $\dfrac{13\pi}{12}\leq x \leq \dfrac{17\pi}{12}$.

The first of these intervals has length $\dfrac{9\pi}{12}$ and the second has length $\dfrac{4\pi}{12}$.

Therefore the fraction of the interval $0\leq x \leq 2\pi$, for which one (or both) of the inequalities is true, is $\dfrac{13\pi}{12}$ out of $2\pi$, which is $\dfrac{13}{24}$. Hence, the solution is (b).