Question

The equation $(\cos p-1)x^2+\left(\cos p\right)x+\sin p=0$ has real roots, then 'P' can take any value in the interval

Answer 1

$(\cos p-1)x^2+\left(\cos p\right)x+\left(\sin p\right)=0$

As roots are real, discriminant$\ge 0$

$△\ge 0\Rightarrow b^2-4ac\ge 0$

$\Rightarrow {\cos }^{2}p-4(\cos p-1)\sin p\ge 0$

$\Rightarrow {\cos }^{2}p-4\cos p\sin p+4\sin p\ge 0$

$\Rightarrow {\cos }^{2}p-4\cos p\sin p+4{\sin }^{2}p-4{\sin }^{2}p+4\sin p$ by adding and subtracting $4{\sin }^{2}p$

$\Rightarrow {(\cos p-2\sin p)}^2+4\sin p(1-\sin p)\ge 0$

$\Rightarrow \cos p-2\sin p$ is always positive.

As $-1\le \sin p\le 1$

$\Rightarrow 1\ge -\sin p\ge -1$

$\Rightarrow 1+1\ge 1-\sin p\ge 0$ by adding $1$

$\Rightarrow 2\ge 1-\sin p\ge 0$

As $\sin p\ge 0$ $\sin p$ is positive and is in first and third quadrant.

$\therefore p\in (0,\pi )$ in first and second quadrant in anticlockwise direction.