Question

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

• A. $\theta =n\pi +{(-1)}^n\alpha$, $\theta =n\pi +{(-1)}^n\beta$
• B. $\theta =n\pi +{(-1)}^n\alpha$, $\theta =n\pi +{(-1)}^{n}2\beta$
• C. $\theta =n\pi +{(-1)}^n\alpha$, $\theta =n\pi +{(-1)}^{n}3\beta$
• D. $\theta =n\pi +{(-1)}^n\alpha$, $\theta =n\pi +{(-1)}^{n}4\beta$

Answer 1

The correct option is A $\theta =n\pi +{(-1)}^n\alpha$, $\theta =n\pi +{(-1)}^n\beta$

We have $4\cos \theta -3\sec \theta =\tan \theta$

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

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

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

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

$\sin \theta =\frac{-1\pm \sqrt{1+16}}{8}$

$=\frac{-1\pm \sqrt{17}}{8}$

$=\frac{-1+\sqrt{17}}{8}$ or $=\frac{-1-\sqrt{17}}{8}$

Now, $\sin \theta =\frac{-1+\sqrt{17}}{8}$. Thus,

$\sin \theta =\sin \alpha$, where $\sin \alpha =\frac{-1+\sqrt{17}}{8}$

$\Rightarrow$ $\theta =n\pi +{(-1)}^n\alpha$, where
$\sin \alpha =\frac{-1+\sqrt{17}}{8}$ and $nϵZ$
and $\sin \theta =\frac{-1-\sqrt{17}}{8}$

$\Rightarrow$ $\sin \theta =\sin \beta$, where $\sin \beta =\frac{-1-\sqrt{17}}{8}$

$\Rightarrow$ $\theta =n\pi +{(-1)}^n\beta$, where

$\sin \beta =\frac{-1-\sqrt{17}}{8}$