Question

If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.

Answer 1

Given: $\cos x=k$
If $k=0$, then
$$\begin{gathered} \cos x=0 \\ \Rightarrow \cos x=\cos \frac{\mathrm{\pi }}{2} \\ \Rightarrow x=(2n+1)\frac{\mathrm{\pi }}{2},n\in Z \end{gathered}$$

Now, $x=\frac{3\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{2},\frac{7\mathrm{\pi }}{2},...$ for $n=1,2,3,...$

If $k=1,$ then

$$\begin{gathered} c\mathrm{os}x=1 \\ \Rightarrow \cos x=\cos 0 \\ \Rightarrow x=2m\mathrm{\pi },\mathrm{m}\in \mathrm{Z} \end{gathered}$$

Now, $x=2\mathrm{\pi },4\mathrm{\pi },6\mathrm{\pi },8\mathrm{\pi },...$ for $m=1,2,3,4,...$

If $k=-1,$ then

$$\begin{gathered} \cos x=-1 \\ \Rightarrow \cos x=\cos \mathrm{\pi } \\ \Rightarrow x=2\mathrm{p\pi }\pm \mathrm{\pi },\mathrm{p}\in \mathrm{Z} \end{gathered}$$

Now,
$x=2p\mathrm{\pi }+\mathrm{\pi },\mathrm{i}.\mathrm{e}.,x=3\mathrm{\pi },5\mathrm{\pi },7\mathrm{\pi },...$ when $p=1,2,3,...$
And,
$x=2p\mathrm{\pi }-\mathrm{\pi },\mathrm{i}.\mathrm{e}.,\mathit{}\mathit{}x=\mathrm{\pi },3\mathrm{\pi },5\mathrm{\pi },7\mathrm{\pi },...$ when $p=1,2,3,4,...$

Clearly, we can see that for $x=\mathrm{\pi }$, $\cos x=k$ has exactly one solution.
∴ $k=-1$