Negative De Moivre

$\huge{ \because (cos0-isin0)=1\\ \therefore 2=2(cos0-isin0)\rightarrow \boxed { 1 } \\ \because -\sin { \theta =\sin { -\theta } } and\cos { \theta } =\cos { -\theta } \\ \therefore (\cos { \frac { \pi }{ 6 } } -i\sin { \frac { \pi }{ 6 } ) } \\ =(\cos { -\frac { \pi }{ 6 } } +i\sin { -\frac { \pi }{ 6 } ) } \\ =(\cos { (-\frac { \pi }{ 6 } +2\pi } )+i\sin { (-\frac { \pi }{ 6 } +2\pi ) } )\\ =(\cos { \frac { 11\pi }{ 6 } } +i\sin { \frac { 11\pi }{ 6 } } )\rightarrow \boxed { 2 } \\ from\quad \boxed { 1 } \quad and\quad \boxed { 2 } \\ { x=\frac { 2 }{ \cos { \frac { \pi }{ 6 } } -i\sin { \frac { \pi }{ 6 } } } }\\ =\frac { 2(cos0-isin0) }{ (\cos { \frac { 11\pi }{ 6 } } +i\sin { \frac { 11\pi }{ 6 } } ) } \\ =\frac { 2 }{ 1 } (\cos { (0-\frac { 11\pi }{ 6 } ) } +i\sin { (0-\frac { 11\pi }{ 6 } ) } )\\ =2(\cos { (-\frac { 11\pi }{ 6 } ) } +i\sin { (-\frac { 11\pi }{ 6 } ) } )\\ =\boxed { \sqrt { 3 } +i } }$

It is known that $e^{i\theta} = \cos{\theta}+i\sin{\theta}$ .

Therefore,

$x = \dfrac {2}{\cos{\frac{\pi}{6}} - i\sin{\frac{\pi}{6}}} = \dfrac {2}{e^{-i\frac{\pi}{6}}} = 2e^{i\frac{\pi}{6}} = 2(\cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}})$

$\quad = 2\left( \frac{\sqrt{3}}{2}+i\frac{1}{2} \right) = \boxed{\sqrt{3}+i}$

$\dfrac{2}{\cos\frac{\pi}{6} - i\sin\frac{\pi}{6}}$ $=\dfrac{2}{\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})}$ $=\dfrac{2}{cis(-\frac{\pi}{6})}$ $=2(cis(-\frac{\pi}{6})^{-1})$ $=2cis(\frac{\pi}{6})=\sqrt{3}+i$