18. Find the sum of all $x$ such that $0 \leq x \leq 360$ and $\cos 12 x^{\circ}=5 \sin 3 x^{\circ}+9 \tan ^{2} x^{\circ}+\cot ^{2} x^{\circ}$.
(2 marks)求所有滿足 $0 \leq x \leq 360$ 及 $\cos 12 x^{\circ}=5 \sin 3 x^{\circ}+9 \tan ^{2} x^{\circ}+\cot ^{2} x^{\circ}$ 的 $x$ 值之和。
18. 540
18. By the AM-GM inequality, $9 \tan ^{2} x^{\circ}+\cot ^{2} x^{\circ} \geq 2 \sqrt{\left(9 \tan ^{2} x^{\circ}\right)\left(\cot ^{2} x^{\circ}\right)}=6$. It follows that the minimum value of the right hand side is 1 . On the other hand, the maximum value of the left hand side is 1 . For equality to hold, both sides must be equal to 1 , and we must have $9 \tan ^{2} x^{\circ}=\cot ^{2} x^{\circ}$ (which implies $\left.\tan x^{\circ}= \pm \frac{1}{\sqrt{3}}\right), \cos 12 x^{\circ}=1$ and $\sin 3 x^{\circ}=-1$.
For $\tan x^{\circ}= \pm \frac{1}{\sqrt{3}}$, the solutions are $x=30,150,210,330$.
For $\cos 12 x^{\circ}=1$, the solutions are $x=0,30,60, \ldots, 330,360$.
For $\sin 3 x^{\circ}=-1$, the solutions are $x=90,210,330$.
Therefore the equation has solutions $x=210,330$ and so the answer is $210+330=540$.