(a) 证明恒等式

(b) 由此求方程

的精确解，其中 $0\leq\theta\leq\pi$。

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。

左边通分后，交叉项相消。再同除以余弦平方，得到正切形式。

On the common domain,

也可以从右边出发。将分母分解成平方差，再把分子拆成与两个因式对应的和。

Write $s=\sin\theta$ and $c=\cos\theta$. Multiplying the numerator and denominator on the right by $c^2$ gives

on the common domain.

用 (a) 化成正切平方的方程。另检查正切无定义但原式有定义的角。

At $\theta=\pi/2$ the original left-hand side is $1$, so this angle is excluded. For other angles in the original domain, (a) gives

hence $\tan\theta=\pm\sqrt3$. The exact solutions on $[0,\pi]$ are $\theta=\pi/3$ and $2\pi/3$. Neither makes an original denominator zero.

也可以用 (a) 通分后的正弦、余弦形式。把分母写成只含正弦平方的式子。

The original left-hand side is $1/(2\sin^2\theta-1)$ wherever defined. Equating it to $2$ gives $2\sin^2\theta-1=1/2$, so $\sin^2\theta=3/4$. Sine is nonnegative on $[0,\pi]$, so $\sin\theta=\sqrt3/2$. This gives $\pi/3$ and $2\pi/3$, both in the original domain.