Question

The equation $(cosp-1)x^2+\left(cosp\right)x+sinp=0$ in the variable x has real roots. Then p can take any value in the interval

• A. $(0,2\pi )$
  
• B. $(-\pi ,0)$
  
• C. $(-\frac{\pi }{2},\frac{\pi }{2})$
  
• D. $(0,\pi )$

Answer 1

The correct option is D $(0,\pi )$

The given equation is
$(cosp-1)x^2+\left(cosp\right)x+sinp=0$
For this equation to have real roots, $D\ge 0$. Thus,
$co{s}^{2}p-4sinp(cosp-1)\ge 0$
or $co{s}^{2}p-4sinpcosp+4si{n}^{2}p+4sinp-4si{n}^{2}p\ge 0$
or $(cosp-2sinp{)}^2+4sinp(1-sinp)\ge 0$
For every real value of p, we have
$(cosp-2sinp{)}^2\ge 0$ and $sinp(1-sinp)\ge 0$
$\therefore D\ge 0,\forall sinp\in (0,\pi )$