Question

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

Answer 1

General soln: $4\cos \theta -3\sec \theta =2\tan \theta$

We are given

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

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

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

$\therefore 4{\sin }^2\theta +2\theta -1=0.....\left(1\right)$

Now, let compare equation $\left(1\right)$ with

$a{x}^2+bx+c=0$

we get $a=4$ & $b=1$ & $c=-1$

now, Discriminant $D=b^2-4ac$

$=(2{)}^2-4(4\left)\right(-1)$

$=4+16$

$D=20>0$

$x=\sin \theta =\frac{-b\pm \sqrt{D}}{2a}$

$\sin \theta =\frac{-2\pm \sqrt{20}}{2\left(4\right)}=\frac{-2\pm \sqrt{20}}{8}$

now,

$\sin \theta =\frac{-2+\sqrt{20}}{8}$ or $\sin \theta =\frac{-2-\sqrt{20}}{8}$

$\sin \theta =\frac{-2+2\sqrt{5}}{8}$ or $\sin \theta =\frac{-2-2\sqrt{5}}{8}$

$\sin \theta =\frac{-1+\sqrt{5}}{4}$ or $\sin \theta =\frac{-1-\sqrt{5}}{8}$

here $\sin {\alpha }_1=\frac{-1+\sqrt{5}}{4},\sin {\alpha }_2=\frac{-1-\sqrt{5}}{4};[-\frac{\pi }{2}\le {\alpha }_1,{\alpha }_2\le \frac{\pi }{2}]$

now, ${\alpha }_1=\frac{\pi }{10}$ & ${\alpha }_2=\frac{-3\pi }{10}$

$\therefore \theta =n\pi +(-1{)}^n\frac{\pi }{10}n\in z$ or $\theta =n\pi +(-1{)}^{\pi }\left(\frac{-3\pi }{10}\right),n\in z$