Review question

# How many solutions does $\cos (\sin x)=\frac{1}{2}$ have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8386

## Solution

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

1. no solutions;

2. one solution;

3. two solutions;

4. three solutions.

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

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

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.