Hint: put $t=cosx$ so we get ${{\cos }^{2k}}\frac{x}{2}\le \frac{1}{2}$

The only real number $k$ which satisfies the desired condition is $k = \frac12.$ First, we will show that $k = \frac12$ actually satisfies the desired condition. We have that when $t \in (-1, 1)$, $(1+t)^{\frac12} (1-t)^{\frac12} = \sqrt{1-t^2} \le 1.$ Hence, it suffices only to show that $k = \frac12$ is the only real number satisfying this condition. Suppose $k \neq \frac12.$ There is then a sufficiently large constant $C$ so that for all $\epsilon \in \mathbb{R}$ which are sufficiently small in absolute, $|(1+\epsilon)^k - (1 + k \epsilon)| < C \epsilon^2$. Therefore, taking $t$ to be sufficiently small gives that $(1+t)^k (1-t)^{1-k} \approx (1+tk - (1-k)t).$ When $k \neq \frac12$, it's easy to get this to be greater than $1.$ $\square$

For $t>0$ and $k\geq 1$ we get $(1+t)^k>1$ and $(1-t)^{1-k}=\frac{1}{(1-t)^{k-1}}>1$, contradiction. Analogously we eliminate $k\leq 0$ (via $t<0$). For $k\in(0,1)$ let us set $t=2k-1\in(-1,1)$ -- then we have $k^k(1-k)^{1-k}\leq \frac{1}{2}$. We will however show that $k^k(1-k)^{1-k}\geq\frac{1}{2}$ for $k\in(0,1)$, and that equality holds only for $k=\frac{1}{2}$. Taking logarithms yields the equivalent \begin{align*} f(k) = k\ln k + (1-k)\ln (1-k) \geq - \ln 2. \end{align*}We have $f'(k) = \ln k - \ln(1-k) = \ln\frac{k}{1-k}$ and so (in $(0,1)$) $f' > 0$ for $k>\frac{1}{2}$, $f'<0$ for $k<\frac{1}{2}$ and $f'=0$ for $k=\frac{1}{2}$. Moreover, $f(\frac{1}{2}) = -\ln 2$. The desired inequality (together with the equality case) follows. Therefore for $k\neq \frac{1}{2}$ the given inequality is not true for all $t$. Conversely, for $k=\frac{1}{2}$ we reach the obvious $\sqrt{1-t^2}\leq1$ for $t\in(-1,1)$.