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Q.
The number of solutions to $\cos (\sin x)=\sin (\cos x), x \in[0,20 \pi]$, is
Trigonometric Functions
Solution:
$\cos (\sin x)=\sin (\cos x)$
$\Rightarrow \cos (\sin x)=\cos (\pi / 2-\cos x)$
$\Rightarrow \sin x=2 n \pi \pm(\pi / 2-\cos x)$
Taking positive sign, $\sin x=2 n \pi+\pi / 2-\cos x$
or $(\cos x+\sin x)=\frac{1}{2}(4 n+1) \pi$
Now $L.H.S. \in[-\sqrt{2}, \sqrt{2}]$,
hence $-\sqrt{2} \leq \frac{1}{2}(4 n+1) \pi \sqrt{2}$
$\Rightarrow \frac{-2 \sqrt{2}-\pi}{4 \pi} \leq n \leq \frac{2 \sqrt{2}-\pi}{4 \pi}$,
which is not satisfied by any integer $n$.
Similarly, taking negative sign we have $\sin x-\cos x=(4 n-1)\pi / 2$,
which is also not satisfied for any integer $n$.
Hence, there is no solution.