一条曲线的方程为

其中 $0\le x\le\pi$。当 $x=a$ 时，曲线的斜率等于 $-\frac32$。

(a) 证明 $a$ 满足方程 $\sin^2 3a+6\sin3a-3=0$。

(b) 由此求 $a$ 的值，其中 $0\le a\le\pi$。

Answer all questions. Write your answers on the answer booklet. Follow the instructions on the front cover of the answer booklet. If you need additional answer paper, ask the invigilator for a continuation booklet.

You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。在答题册上作答，并遵守答题册封面的指示。如需额外答题纸，请向监考人员索取续答册。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

用商法则求导，先利用 $\sin^2+\cos^2=1$ 简化分子。令梯度为题给值后，余弦的一次项会抵消，再消去余弦平方。

Writing $s=\sin3x$ and $c=\cos3x$, the quotient rule gives

At $x=a$ the gradient is $-3/2$. Since $2-c>0$, cross-multiplication gives

Expanding the right side and cancelling $6c$ gives

At $x=a$, this is $\sin^23a+6\sin3a-3=0$, as required.

先解关于 $\sin3a$ 的二次方程。$3a$ 介于 $0$ 与 $3\pi$，正弦为正的两段都需要取解。

Part (a) gives $\sin3a=-3\pm2\sqrt3$. Only $-3+2\sqrt3$ lies in $[-1,1]$. Let $\gamma=\arcsin(-3+2\sqrt3)$. In $0\le3a\le3\pi$,

Thus the values of $a$, to three significant figures, are