Question

Solve $4\cos \theta -3\sec \theta =2\tan \theta$.

Answer 1

$4\cos \theta -\frac{3}{\cos \theta }=\frac{2\sin \theta }{\cos \theta }$

$4(1-{\sin }^2\theta )=2\sin \theta +3$

$4{\sin }^2\theta +2\sin \theta -1=0$

$\therefore \sin \theta =\frac{-2\pm \sqrt{(4+16)}}{8}$

$=\frac{-2\pm 2\sqrt{5}}{8}$

$=\frac{\sqrt{5}-1}{4},-\left(\frac{\sqrt{5}+1}{4}\right)$

$\sin \theta =\sin {18}^o$ i.e, $\sin \left(\pi /10\right)$,

$\therefore \theta =n\pi +(-1{)}^n(\pi /10)$

$\sin \theta =-\cos {36}^o=-\sin {54}^o$

$=\sin (-3\pi /10)$

$\therefore \theta =n\pi -(-1{)}^n(3\pi /10)$.