Question

The equation $(Cosp-1)x^2+\left(Cosp\right)x+Sinp=0,$ where x is a variable with real roots. then the interval of p may be any one of the following.

• 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 equation has real roots if
$co{s}^{2}p-4(cosp-1)sinp\ge 0$
Or $co{s}^{2}p-4cosp.sinp+4sinp\ge 0$
Or $(cosp-2sinp{)}^2-4si{n}^{2}p+4sinp\ge 0$...(1)
$(cosp-2sinp{)}^2+4sinp(1-sinp)\ge 0$
Or $(cosp-2sinp{)}^2\ge 0.1,1-sinp\ge 0$ for all values of p and for $pϵ[0,\pi ]$
$sinp\ge 0$ so that the discriminant is non-negative.
The correct answer is: $(0,\pi )$.