题目
Problem
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(i) Solve, for , the equation
3\sin x\tan x=11+\cos x \end{align*}$$ giving the answers in radians to 3 decimal places. <div style="text-align: right;">(5)</div> (ii) Given that - $0<\theta<90°$ - $\cos\theta=\dfrac{1}{3}$ find, in simplest form, the exact value of $\tan\theta$ <div style="text-align: right;">(2)</div>解答
(i)
解法一
思路
展开
把 写成 ，再用 把整个方程化成关于 的二次方程。
答题过程
展开
Writing ,
3\sin x\cdot\frac{\sin x}{\cos x}=11+\cos x \end{align*}$$ $$\begin{align*} \frac{3\sin^2x}{\cos x}=11+\cos x. \end{align*}$$ Multiplying both sides by $\cos x$ (noting $\cos x\neq0$): $$\begin{align*} 3\sin^2x=11\cos x+\cos^2x. \end{align*}$$ Using $\sin^2x=1-\cos^2x$: $$\begin{align*} 3(1-\cos^2x)=11\cos x+\cos^2x \end{align*}$$ $$\begin{align*} 3-3\cos^2x=11\cos x+\cos^2x \end{align*}$$ $$\begin{align*} 4\cos^2x+11\cos x-3=0. \end{align*}$$ Factorising: $$\begin{align*} (4\cos x-1)(\cos x+3)=0. \end{align*}$$ Since $-1\leqslant\cos x\leqslant1$, the factor $\cos x+3=0$ has no solution. So $$\begin{align*} \cos x=\frac{1}{4}. \end{align*}$$ $$\begin{align*} x=\arccos\!\left(\frac{1}{4}\right)\approx1.318\quad\text{or}\quad x=2\pi-\arccos\!\left(\frac{1}{4}\right)\approx4.965. \end{align*}$$ $$\begin{align*} \boxed{x\approx1.318\text{ or }x\approx4.965} \end{align*}$$ </details> ## (ii) ### 解法一 #### 思路 <details> <summary>展开</summary> 已知 $\cos\theta=\frac{1}{3}$，用 $\sin^2\theta+\cos^2\theta=1$ 求 $\sin\theta$，再算 $\tan\theta=\frac{\sin\theta}{\cos\theta}$。 </details> #### 答题过程 <details> <summary>展开</summary> Since $\cos\theta=\dfrac{1}{3}$ and $0<\theta<90°$, $$\begin{align*} \sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}. \end{align*}$$ Therefore $$\begin{align*} \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\;\dfrac{2\sqrt{2}}{3}\;}{\;\dfrac{1}{3}\;}=2\sqrt{2}. \end{align*}$$ </details>