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

\[\cos x = 0\]

\[ \Rightarrow \cos x = \cos \frac{\pi}{2}\]

\[ \Rightarrow x = (2n + 1) \frac{\pi}{2}, n \in Z\]
Now,

\[x = \frac{3\pi}{2} , \frac{5\pi}{2}, \frac{7\pi}{2}, . . .\]  for 

n = 1, 2, 3, . . .
If k = 1, then

\[cos x = 1\]

\[ \Rightarrow \cos x = \cos 0\]

\[ \Rightarrow x = 2m\pi, m \in Z\]
Now, \[x = 2\pi, 4\pi, 6\pi, 8\pi, . . .\]

\[m = 1, 2, 3, 4, . . .\]
If \[k = - 1,\] then

\[\cos x = - 1\]

\[ \Rightarrow \cos x = \cos \pi\]

\[ \Rightarrow x = 2p\pi \pm \pi, p \in Z\]
Now,

\[x = 2p\pi + \pi, i . e . , x = 3\pi, 5\pi, 7\pi, . . .\] when
p = 1, 2, 3, . . .
And \[x = 2p\pi - \pi, i . e . , x = \pi, 3\pi, 5\pi, 7\pi, . . .\] when

p = 1, 2, 3, 4, . . .
Clearly, we can see that for \[x = \pi\]

\[\cos x = k\] has exactly one solution.
∴ k = - 1