How many solutions does $\cos (\sin x)=\frac{1}{2}$ have?

How many solutions does $\cos (\sin x)=\frac{1}{2}$ have?

Question

In the range $0 \le x < 2\pi$ the equation $\cos(\sin x) = \frac{1}{2}$ has

1. no solutions;

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1. one solution;

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1. two solutions;

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1. three solutions.

Suggestion

First consider $\cos y = \frac{1}{2};$ what are the solutions of this equation?

From this, can you find the values of $\sin x$ that satisfy $\cos(\sin x) = \frac{1}{2}?$

Do all or any of these values of $\sin x$ give a solution for $x$?

Solution

The equation $\cos y = \dfrac{1}{2}$ has solutions $y = \pm \frac{\pi}{3}, \pm \frac{5\pi}{3}, \pm \frac{7\pi}{3}, \pm \frac{11\pi}{3}, \dots c .$

So if $\cos(\sin x) = \dfrac{1}{2}$ then $\sin x = \pm \frac{\pi}{3}, \pm \frac{5\pi}{3}, \pm \frac{7\pi}{3}, \pm \frac{11\pi}{3}, \dots c .$

Now as $\pi > 3$, we have $\dfrac{\pi}{3} > 1$.

Therefore, all numbers on this list have magnitude greater than $1$, and so lie outside the range of sine ($-1 \le \sin x \le 1$).

This means that $\sin x$ cannot equal any of the numbers on this list, so there are no solutions.

The answer is (a).