Find the range of

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

Step 1: Simplify the inner trigonometric expression

Let the expression inside the logarithm be $$g(x)$$.

Group the terms to apply sum and difference formulas:

$$g(x) = 3 + \left[ \cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) \right] + \left[ \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) \right]$$

Use the standard identities
$$\cos(A+B) - \cos(A-B) = -2\sin A \sin B$$ and
$$\cos(A+B) + \cos(A-B) = 2\cos A \cos B$$:

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

$$g(x) = 3 - 2\left(\frac{1}{\sqrt{2}}\right)\sin(x) + 2\left(\frac{1}{\sqrt{2}}\right)\cos(x)$$

$$g(x) = 3 - \sqrt{2}\sin x + \sqrt{2}\cos x$$

Step 2: Find the range of $$g(x)$$

An expression in the form $$a\sin x + b\cos x$$ has a range of $$[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$$.

For our expression $$-\sqrt{2}\sin x + \sqrt{2}\cos x$$, the range is:

$$[-\sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2}, \sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2}]$$

$$[-\sqrt{2+2}, \sqrt{2+2}]$$

$$[-2, 2]$$

Since $$g(x) = 3 + (-\sqrt{2}\sin x + \sqrt{2}\cos x)$$, we add 3 to this range:

Range of $$g(x) = [3 - 2, 3 + 2] = [1, 5]$$

Step 3: Find the range of the logarithmic function

The original function is $$f(x) = \log_{\sqrt{5}}(g(x))$$.

Since the base $$\sqrt{5}$$ is greater than 1, the logarithm is an increasing function. We can find the range of $$f(x)$$ by evaluating it at the boundaries of the range of $$g(x)$$.

Minimum value = $$\log_{\sqrt{5}}(1) = 0$$

Maximum value = $$\log_{\sqrt{5}}(5) = 2$$

The range of $$f(x)$$ is $$[0, 2]$$.