(a) 证明方程

可写成 $a\sin^2\theta+b\sin\theta+c=0$ 的形式，其中 $a$、$b$ 和 $c$ 是需要求出的常数。

(b) 由此解方程

其中 $0^\circ\leq\theta\leq360^\circ$。

Answer all questions. 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。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。

通分后用平方差处理分母，再把余弦平方写成正弦平方。

On the domain of the original equation, the common denominator is nonzero and the equation becomes

Thus $2\sin\theta=2\sin^2\theta-1$, or $2\sin^2\theta-2\sin\theta-1=0$. One choice is $(a,b,c)=(2,-2,-1)$.

用求根公式求两个正弦值，舍去大于 $1$ 的根，再找第三、第四象限的角。

Using the quadratic formula on the equation in (a),

Only $(1-\sqrt3)/2$ lies in $[-1,1]$. The reference angle is $\arcsin((\sqrt3-1)/2)\approx21.4707^\circ$, so the solutions are $201.5^\circ$ and $338.5^\circ$. Neither makes an original denominator zero.