Question

If $(cose{c}^2\theta -4)x^2+(cot\theta +\sqrt{3})x+co{s}^2\frac{3\pi }{2}=0$ holds true for all real x, then the most general values of $\theta$ can be given by $(n\in z)$

• A. $2n\pi +\frac{11\pi }{6}$
• B. $2n\pi +\frac{5\pi }{6}$
• C. $2n\pi -\frac{7\pi }{6}$
• D. $n\pi \pm \frac{11\pi }{6}$

Answer 1

Answer: (A), (C)

The correct options are
A $2n\pi +\frac{11\pi }{6}$
C $2n\pi -\frac{7\pi }{6}$

We have, $(cose{c}^2\theta -4)x^2+(cot\theta +\sqrt{3}).x+co{s}^2\frac{3\pi }{2}=0$
$(cose{c}^2\theta -4)x^2+(cot\theta +\sqrt{3}).x=0$

$x\left[\right(cose{c}^2\theta -4)x^2+(cot\theta +\sqrt{3}\left)\right]=0$.

Hence we get one root as $x=0$.

Now since the equation is true for all real x,
$B=0$

$cot\theta =-\sqrt{3}$.

$\theta =\pi -\frac{\pi }{6}$
and
$\theta =2\pi -\frac{\pi }{6}$.

Hence
$\theta \in n\pi -\frac{\pi }{6}$. where $n\in N$.

$\theta =2n\pi +\frac{11\pi }{6}$ and $\theta =2n\pi -\frac{7\pi }{6}$.