Question

Find the most general value of $\theta$ satisfying the equation tan $\theta =-1$ and $\cos \theta =\frac{1}{\sqrt{2}}$

Answer 1

Given:
$\tan \theta =-1$ and $\cos \theta =\frac{1}{\sqrt{2}}$
Period of $\cos \theta =2\pi$ and
period of $\tan \theta =\pi$
$\therefore$ Common period
$=LCM$ of $2\pi$ and $\pi =2\pi$
Since one common period range $=[0,2\pi ]$
$\tan \theta =-1\Rightarrow \theta =\frac{3\pi }{4},\frac{7\pi }{4}$
and $\cos \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4},\frac{7\pi }{4}$
$\therefore$ Common value of $\theta =\frac{7\pi }{4}$
General solution will be
$\theta =n$ (common period) $+$ common value
$\Rightarrow \theta =2n\pi +\frac{7\pi }{4},nϵZ$