Question

The general solutions of $\tan 2x=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$ is

• A. $x=n\pi +\frac{\pi }{10},n\in Z$
• B. $x=\frac{n\pi }{2}+\frac{\pi }{20},n\in Z$
• C. $x=\frac{n\pi }{2}+\frac{\pi }{10},n\in Z$
• D. $x=\frac{n\pi }{4}+\frac{\pi }{20},n\in Z$

Answer 1

The correct option is B $x=\frac{n\pi }{2}+\frac{\pi }{20},n\in Z$

$\tan 2x=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}=\tan {18}^{\circ }\Rightarrow \tan 2x=\tan \frac{\pi }{10}\Rightarrow 2x=n\pi +\frac{\pi }{10},n\in Z\therefore x=\frac{n\pi }{2}+\frac{\pi }{20},n\in Z$