The range of the function f(x) = log_{sqrt{5} | vedclass

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(D) Let the argument of the logarithm be $g(x) = 3 + \cos(\frac{3\pi}{4} + x) + \cos(\frac{\pi}{4} + x) + \cos(\frac{\pi}{4} - x) - \cos(\frac{3\pi}{4

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$[0, 2]$

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(D) Let the argument of the logarithm be $g(x) = 3 + \cos(\frac{3\pi}{4} + x) + \cos(\frac{\pi}{4} + x) + \cos(\frac{\pi}{4} - x) - \cos(\frac{3\pi}{4} - x)$.Using the sum-to-product formulas $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$ and $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$:$g(x) = 3 + [\cos(\frac{3\pi}{4} + x) - \cos(\frac{3\pi}{4} - x)] + [\cos(\frac{\pi}{4} + x) + \cos(\frac{\pi}{4} - x)]$$g(x) = 3 - 2\sin(\frac{3\pi}{4})\sin(x) + 2\cos(\frac{\pi}{4})\cos(x)$Since $\sin(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:$g(x) = 3 - 2(\frac{1}{\sqrt{2}})\sin(x) + 2(\frac{1}{\sqrt{2}})\cos(x) = 3 + \sqrt{2}(\cos x - \sin x)$.We know that $-\sqrt{2} \leq \cos x - \sin x \leq \sqrt{2}$.Thus,$3 + \sqrt{2}(-\sqrt{2}) \leq g(x) \leq 3 + \sqrt{2}(\sqrt{2})$,which simplifies to $3 - 2 \leq g(x) \leq 3 + 2$,or $1 \leq g(x) \leq 5$.Since $f(x) = \log_{\sqrt{5}}(g(x))$,the range is $[\log_{\sqrt{5}}(1), \log_{\sqrt{5}}(5)]$.Since $\log_{\sqrt{5}}(1) = 0$ and $\log_{\sqrt{5}}(5) = \frac{\log_5(5)}{\log_5(\sqrt{5})} = \frac{1}{1/2} = 2$,the range is $[0, 2]$.

The range of the function $f(x) = \log_{\sqrt{5}}(3 + \cos(\frac{3\pi}{4} + x) + \cos(\frac{\pi}{4} + x) + \cos(\frac{\pi}{4} - x) - \cos(\frac{3\pi}{4} - x))$ is:

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