\[k = 1: x\in \mathbb{R}\\\\ k = 3: cos(3x)=4cos(x) \Rightarrow cos^3(x)-3cos(x)sin^2(x)=4cos(x)\\\\\Rightarrow cos^3(x)=\frac{7}{4}cos(x) \\\]
- which only has the trivial solution: $cos(x)=0,\;\; i.e. \;\;x =\frac{\pi }{2}+p\pi,\;\;p\in \mathbb{Z}$
\[k = 5,7,9 ...: cos(kx)=2^{k-3}cos(3x)=2^{k-3}4cos(x)\]
So I end up with the same solution as for $k =3$.
I guess there are other (better) ways to solve the equation. Maybe one could use the inequality:
\[\left | \frac{cos(kx)}{kcos(x)} \right |\leq 1\]
- to show that the only non-trivial solution ($cos(x)\neq 0$) appears for $k=1$
Another way could be to define the function:
\[f(x)=cos(kx)-2^{k-1}cos(x), \;\; k = 3,5,7 ...\]
I would look at the set: $S = \left \{ x\in\mathbb{R}|f(x)=0 \right \}$
Now, $f'(x)=-ksin(kx)+2^{k-1}sin(x)$ so there is at least one inflection point between $0$ and $\pi$
But $f''(x)=-k^2cos(kx)+2^{k-1}cos(x)=(1-k^2)cos(kx)$ for $x\in S$
$f’’(x)=0$ is only possible if $k = 1$ or $x=\frac{\pi}{2}\in S$ and due to periodicity $S = \left \{x|x=\frac{\pi}{2}+p\pi, p\in\mathbb{Z} \right \}$
Thankyou for any corrections to the above considerations!
The calculus method that you showed is great, I think, because I don't really go through them, (I have a little headache today)...but your proposed method suggests an easier way to approach the problem as I see if:
$\left| \frac{cos(kx)}{kcos(x)} \right|\le 1$ we have $\left| kcos(x) \right|\ge \left| cos(kx) \right|$ but since $\cos kx =2^{k-1} \cos x$ and $k>0$, we get
$k \left| cos(x) \right|\ge 2^{k-1}\left| \cos x \right|$
This is like asking what $k$ will it hold so that $k \ge 2^{k-1}$ and obviously that inequality is true iff $k=1$.
And the result follows where we have $x=\frac{\pi}{2}+p\pi, p\in\mathbb{Z} $.
The calculus method that you showed is great, I think, because I don't really go through them, (I have a little headache today)...but your proposed method suggests an easier way to approach the problem as I see if:
$\left| \frac{cos(kx)}{kcos(x)} \right|\le 1$ we have $\left| kcos(x) \right|\ge \left| cos(kx) \right|$ but since $\cos kx =2^{k-1} \cos x$ and $k>0$, we get
$k \left| cos(x) \right|\ge 2^{k-1}\left| \cos x \right|$
This is like asking what $k$ will it hold so that $k \ge 2^{k-1}$ and obviously that inequality is true iff $k=1$.
And the result follows where we have $x=\frac{\pi}{2}+p\pi, p\in\mathbb{Z} $.
I see 2 cases
1) k = 1 for which is it is identity
or 2) cos kx =0 and cos x =0 or which $x=\frac{\pi}{2}+p\pi, p\in\mathbb{Z} $
)...but your proposed method suggests an easier way to approach the problem as I see if: