1. Simplify the trigonometric expression inside the logarithm

We need the range of

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

Let

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

Now use

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

Also,

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

Hence,

$$S=3+\sqrt2\cos x-\sqrt2\sin x.$$

Now write

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

So,

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

Therefore,

$$S=3+2\cos\left(x+\tfrac{\pi}{4}\right).$$

2. Find the range of the logarithm argument

Since

$$-1\le \cos\left(x+\tfrac{\pi}{4}\right)\le 1,$$

we get

$$3-2\le S\le 3+2.$$

Thus,

$$1\le S\le 5.$$

So the argument of the logarithm varies in

$$[1,5].$$

This is valid since the logarithm argument is always positive.

3. Find the range of the logarithmic function

Now

$$f(x)=\log_{\sqrt5} S, \qquad S\in[1,5].$$

Since the base $\sqrt5>1$, the logarithm is increasing. Therefore,

$$\min f(x)=\log_{\sqrt5}(1)=0,$$ $$\max f(x)=\log_{\sqrt5}(5).$$

Now,

$$\log_{\sqrt5}(5)=2$$

because

$$(\sqrt5)^2=5.$$

Hence the range is

$$[0,2].$$

4. Check options

• A: $(0,\sqrt5)$ — incorrect
• B: $[-2,2]$ — incorrect
• C: $\left[\frac1{\sqrt5},\sqrt5\right]$ — incorrect
• D: $[0,2]$ — correct

Therefore, the correct answer is D.