(a) 证明方程

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

(b) 由此解方程 $5\cos\theta-\sin\theta\tan\theta+1=0$，其中 $0<\theta<2\pi$。

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。

用正弦与余弦的比值表示正切，乘以余弦后把正弦平方改写成余弦平方。

Where $\cos\theta\ne0$, multiplying the original equation by $\cos\theta$ gives

Using $\sin^2\theta=1-\cos^2\theta$ gives $6\cos^2\theta+\cos\theta-1=0$. Thus $(a,b,c)=(6,1,-1)$.

因式分解得到两个余弦值。每个值在一周内各对应两个角。

The quadratic factors as $(3\cos\theta-1)(2\cos\theta+1)=0$, giving $\cos\theta=1/3$ or $-1/2$. The first gives $\cos^{-1}(1/3)$ and $2\pi-\cos^{-1}(1/3)$, and the second gives $2\pi/3$ and $4\pi/3$. All are in the domain of the original equation. In increasing order they are $1.23$, $2.09$, $4.19$ and $5.05$ radians to three significant figures.