The equation (cos p - 1) x^2 + (cos p) x + si | vedclass

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(D) Given the quadratic equation $(\cos p - 1) x^2 + (\cos p) x + \sin p = 0$. For the equation to have real roots,the discriminant $D = b^2 - 4ac \ge

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$(0, \pi)$

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(D) Given the quadratic equation $(\cos p - 1) x^2 + (\cos p) x + \sin p = 0$. For the equation to have real roots,the discriminant $D = b^2 - 4ac \geq 0$. Here,$a = \cos p - 1$,$b = \cos p$,and $c = \sin p$. Substituting these into the discriminant condition: $(\cos p)^2 - 4(\cos p - 1)(\sin p) \geq 0$ $\cos^2 p - 4\sin p \cos p + 4\sin p \geq 0$ Since $\cos p - 1 
eq 0$,we have $\cos p 
eq 1$,which implies $p 
eq 2n\pi$. For $p \in (0, \pi)$,$\sin p > 0$ and $\cos p - 1 The condition simplifies to $p \in (0, \pi)$ as it satisfies the discriminant inequality for real roots.

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

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