Overcomplicated Log Function

In order to maximize $f(x)$ we need to minimize $g(x) = \log_{\sqrt{2}}(16\sin^{2}(x) + 1)$ . Now since $0 \le \sin^{2}(x) \le 1$ we have that the minimum of $(16\sin^{2}(x) + 1)$ is $1$ , giving a minimum for $g(x)$ of $\log_{\sqrt{2}}(1) = 0$ . This in turn gives a maximum for $f(x)$ of $\log_{2}(2) = \boxed{1}$ .