Trigonometry (11th Edition) Clone

The solutions are $$x=\{\frac{\pi}{18},\frac{7\pi}{18},\frac{13\pi}{18},\frac{19\pi}{18},\frac{25\pi}{18},\frac{31\pi}{18}\}$$
$$\cot3x=\sqrt3$$ over interval $[0,2\pi)$ 1) Interval $[0,2\pi)$ can be written as $$0\le x\lt2\pi$$ That means, for $3x$, the interval would be $$0\le3x\lt6\pi$$ or $$3x\in[0,6\pi)$$ 2) Now consider back the equation $$\cot3x=\sqrt3$$ Over the interval $[0,6\pi)$, there are 6 values whose $\cot$ equals $\sqrt3$, which are $\frac{\pi}{6},\frac{7\pi}{6},\frac{13\pi}{6},\frac{19\pi}{6},\frac{25\pi}{6},\frac{31\pi}{6}$, meaning that $$3x=\{\frac{\pi}{6},\frac{7\pi}{6},\frac{13\pi}{6},\frac{19\pi}{6},\frac{25\pi}{6},\frac{31\pi}{6}\}$$ So $$x=\{\frac{\pi}{18},\frac{7\pi}{18},\frac{13\pi}{18},\frac{19\pi}{18},\frac{25\pi}{18},\frac{31\pi}{18}\}$$ (In fact, a value whose $\tan$ equals $\frac{\sqrt3}{3}$ would at the same times have $\cot$ equaling $\sqrt3$)