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(A) The point $(a, 0)$ lies on the curve $y = \sin 2x - \sqrt{3} \sin x$. Setting $y = 0$,we get $0 = 2 \sin x \cos x - \sqrt{3} \sin x = \sin x (2 \c

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$4A + 8 \cos a = 7$

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(A) The point $(a, 0)$ lies on the curve $y = \sin 2x - \sqrt{3} \sin x$. Setting $y = 0$,we get $0 = 2 \sin x \cos x - \sqrt{3} \sin x = \sin x (2 \cos x - \sqrt{3})$. Since $a > 0$ is the first point of intersection with the positive $x$-axis,we have $\sin a 
eq 0$,so $2 \cos a = \sqrt{3}$,which gives $a = \frac{\pi}{6}$. The area $A$ is given by the integral $A = \int_{0}^{\pi/6} (\sin 2x - \sqrt{3} \sin x) dx$. Evaluating the integral: $A = \left[ -\frac{\cos 2x}{2} + \sqrt{3} \cos x ight]_{0}^{\pi/6}$. Substituting the limits: $A = \left( -\frac{\cos(\pi/3)}{2} + \sqrt{3} \cos(\pi/6) ight) - \left( -\frac{\cos 0}{2} + \sqrt{3} \cos 0 ight)$. $A = \left( -\frac{1/2}{2} + \sqrt{3} \cdot \frac{\sqrt{3}}{2} ight) - \left( -\frac{1}{2} + \sqrt{3} ight) = \left( -\frac{1}{4} + \frac{3}{2} ight) - \left( -\frac{1}{2} + \sqrt{3} ight) = \frac{5}{4} + \frac{1}{2} - \sqrt{3} = \frac{7}{4} - \sqrt{3}$. Thus,$4A = 7 - 4\sqrt{3}$. Since $a = \pi/6$,$\cos a = \sqrt{3}/2$,so $8 \cos a = 4\sqrt{3}$. Substituting this into the expression,$4A + 8 \cos a = (7 - 4\sqrt{3}) + 4\sqrt{3} = 7$.

If $(a, 0); a > 0$ is the point where the curve $y = \sin 2x - \sqrt{3} \sin x$ cuts the $x$-axis first,and $A$ is the area bounded by this part of the curve,the origin,and the positive $x$-axis,then:

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