If 5 sin theta + 3 cos left(theta + frac{pi}{ | vedclass

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(B) Let $f(	heta) = 5 \sin 	heta + 3 \cos \left(	heta + \frac{\pi}{3}ight) + 3$. Expanding the cosine term: $f(	heta) = 5 \sin 	heta + 3 \left(

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(B) Let $f(	heta) = 5 \sin 	heta + 3 \cos \left(	heta + \frac{\pi}{3}ight) + 3$. Expanding the cosine term: $f(	heta) = 5 \sin 	heta + 3 \left(\cos 	heta \cos \frac{\pi}{3} - \sin 	heta \sin \frac{\pi}{3}ight) + 3$. $f(	heta) = 5 \sin 	heta + 3 \left(\frac{1}{2} \cos 	heta - \frac{\sqrt{3}}{2} \sin 	hetaight) + 3$. $f(	heta) = \left(5 - \frac{3\sqrt{3}}{2}ight) \sin 	heta + \frac{3}{2} \cos 	heta + 3$. This is of the form $A \sin 	heta + B \cos 	heta + C$,where $A = 5 - \frac{3\sqrt{3}}{2}$,$B = \frac{3}{2}$,and $C = 3$. The range of $A \sin 	heta + B \cos 	heta$ is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$. $A^2 + B^2 = \left(5 - \frac{3\sqrt{3}}{2}ight)^2 + \left(\frac{3}{2}ight)^2 = 25 - 15\sqrt{3} + \frac{27}{4} + \frac{9}{4} = 25 - 15\sqrt{3} + 9 = 34 - 15\sqrt{3}$. So the range of $f(	heta)$ is $[3 - \sqrt{34 - 15\sqrt{3}}, 3 + \sqrt{34 - 15\sqrt{3}}]$. Thus,$\alpha = 3 - \sqrt{34 - 15\sqrt{3}}$ and $\beta = 3 + \sqrt{34 - 15\sqrt{3}}$. Then $\alpha + \beta = 6$ and $\alpha - \beta = -2\sqrt{34 - 15\sqrt{3}}$. Substituting into the expression: $(\alpha - \beta)(\alpha + \beta - 6) = (-2\sqrt{34 - 15\sqrt{3}})(6 - 6) = 0$.

If $5 \sin \theta + 3 \cos \left(\theta + \frac{\pi}{3}\right) + 3$ lies between $\alpha$ and $\beta$ (including $\alpha, \beta$ also),then $(\alpha - \beta)(\alpha + \beta - 6) = $

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