Question

General solution of ${\sin }^{2}x-5\sin x\cos x-6{\cos }^{2}x=0$ is

• A. $x=n\pi -\frac{\pi }{4},nϵz$ only
• B. $n\pi +{\tan }^{-1}6,nϵz$ only
• C. Both $a$ and $b$
• D. None of these

Answer 1

The correct option is C Both $a$ and $b$

${\sin }^{2}x-5\sin x\cos x-6{\cos }^{2}x=0$
Dividing by ${\cos }^{2}x$,
${\tan }^{2}x-5\tan x-6=0$
$\tan x=6or-1$
$\tan x=\tan \left(\frac{-\pi }{4}\right)$ or $\tan \left({\tan }^{-1}6\right)$
So,
$x=n\pi -\frac{\pi }{4}$ or $n\pi +{\tan }^{-1}6$