Find the range of

$$f(x)=\log_{\sqrt5}\left(3+\cos\left(\frac{3\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}+x\right)-\cos\left(\frac{\pi}{4}-x\right)-\cos\left(\frac{3\pi}{4}-x\right)\right).$$

Let

$$A=3+\cos\left(\frac{3\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}+x\right)-\cos\left(\frac{\pi}{4}-x\right)-\cos\left(\frac{3\pi}{4}-x\right).$$

Using

$$\cos(a+b)-\cos(a-b)=-2\sin a\sin b,$$

we get

$$\cos\left(\frac{3\pi}{4}+x\right)-\cos\left(\frac{3\pi}{4}-x\right)=-2\sin\frac{3\pi}{4}\sin x=-\sqrt2\sin x.$$

Also,

$$\cos\left(\frac{\pi}{4}+x\right)-\cos\left(\frac{\pi}{4}-x\right)=-2\sin\frac{\pi}{4}\sin x=-\sqrt2\sin x.$$

Hence

$$A=3-2\sqrt2\sin x.$$

Since

$$-1\le \sin x\le 1,$$

we obtain

$$3-2\sqrt2\le A\le 3+2\sqrt2.$$

Now

$$3-2\sqrt2=(\sqrt2-1)^2>0,$$

so the logarithm is defined throughout.

Since the base

$$\sqrt5>1,$$

the logarithmic function is increasing.

Therefore the range of

$$f(x)$$

is

$$\left[\log_{\sqrt5}(3-2\sqrt2),\ \log_{\sqrt5}(3+2\sqrt2)\right].$$

Using

$$3+2\sqrt2=(\sqrt2+1)^2,\qquad 3-2\sqrt2=(\sqrt2-1)^2,$$

the range may also be written as

$$\left[\log_{\sqrt5}(\sqrt2-1)^2,\ \log_{\sqrt5}(\sqrt2+1)^2\right].$$

Therefore, $$\text{Range}=\left[\log_{\sqrt5}(3-2\sqrt2),\ \log_{\sqrt5}(3+2\sqrt2)\right].$$