When the function f(x) = 2(cos 3x + cos sqrt{ | vedclass

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(A) Given the function $f(x) = 2(\cos 3x + \cos \sqrt{3} x)$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}

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(A) Given the function $f(x) = 2(\cos 3x + \cos \sqrt{3} x)$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:$f(x) = 2 	imes 2 \cos(\frac{3 + \sqrt{3}}{2} x) \cos(\frac{3 - \sqrt{3}}{2} x) = 4 \cos(\frac{3 + \sqrt{3}}{2} x) \cos(\frac{3 - \sqrt{3}}{2} x)$.The maximum value of $\cos 	heta$ is $1$.For $f(x)$ to be $4$,both $\cos(\frac{3 + \sqrt{3}}{2} x)$ and $\cos(\frac{3 - \sqrt{3}}{2} x)$ must be equal to $1$ or both must be equal to $-1$.Case $1$: $\cos(\frac{3 + \sqrt{3}}{2} x) = 1$ and $\cos(\frac{3 - \sqrt{3}}{2} x) = 1$.This implies $\frac{3 + \sqrt{3}}{2} x = 2n\pi$ and $\frac{3 - \sqrt{3}}{2} x = 2m\pi$ for integers $n, m$.Dividing the two equations: $\frac{3 + \sqrt{3}}{3 - \sqrt{3}} = \frac{n}{m}$.Rationalizing the denominator: $\frac{(3 + \sqrt{3})^2}{9 - 3} = \frac{9 + 3 + 6\sqrt{3}}{6} = \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3}$.Since $2 + \sqrt{3}$ is irrational,there are no non-zero integers $n, m$ that satisfy this. Thus,$n=0, m=0$ is the only solution,giving $x=0$.Case $2$: Both are $-1$. This leads to a similar contradiction as the ratio of the arguments is irrational.Therefore,the only value of $x$ for which the function attains its maximum value is $x=0$. The number of values is $1$.

When the function $f(x) = 2(\cos 3x + \cos \sqrt{3} x)$ attains its maximum value,what is the number of values of $x$?

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