解方程 $2\cos x-\cos\frac12x=1$，其中 $0\le x\le2\pi$。

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。

把 $\cos x$ 写成关于 $\cos(x/2)$ 的二倍角式，解二次方程。新角范围是 $[0,\pi]$，找到新角后还要乘回 $2$。

Put $u=\cos(x/2)$. Since $\cos x=2\cos^2(x/2)-1$, the equation becomes

Hence

Thus $\cos(x/2)=1$ or $\cos(x/2)=-3/4$.

Because $0\le x/2\le\pi$, the first gives $x=0$ and the second gives

Therefore

in radians.

也可先将半角余弦单独移项，平方后得到关于 $\cos x$ 的二次方程。平方可能产生伪根，必须代回未平方的关系检查。

Rearrange the equation as

Squaring and using $\cos^2(x/2)=(1+\cos x)/2$ gives

Thus

For $\cos x=1$, the candidates in the interval are $x=0,2\pi$. The original relation requires $\cos(x/2)=1$, so $x=0$ is valid and $x=2\pi$ is rejected.

For $\cos x=1/8$, the original relation requires $\cos(x/2)=-3/4$. Thus $x/2$ is in the second quadrant, selecting

Hence $\boxed{x=0\text{ or }x=4.84}$ in radians.