(a) 解方程 $6\sqrt y+\frac2{\sqrt y}-7=0$。

(b) 由此解方程 $6\sqrt{\tan x}+\frac2{\sqrt{\tan x}}-7=0$，其中 $0^\circ\le x\le360^\circ$。

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。

设 $u=\sqrt y$。原式分母要求 $u>0$，再乘以 $u$ 化成二次方程。

Let $u=\sqrt y>0$. Multiplication by $u$ gives

Both roots, $2/3$ and $1/2$, are positive. Hence $y=4/9$ or $1/4$.

把 (a) 的两个 $y$ 值作为正切值。每个正值在第一和第三象限各有一个角。

The equation is the one in (a) with $y=\tan x$. Thus $\tan x=4/9$ or $1/4$. The first gives $24.0^\circ$ and $204.0^\circ$, and the second gives $14.0^\circ$ and $194.0^\circ$, all to one decimal place. All four give a positive, defined tangent.