If the roots of the given equation (cos p - 1 | vedclass

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Question

(C) Given equation: $(\cos p - 1){x^2} + (\cos p)x + \sin p = 0$. Since the roots are real,the discriminant $D \ge 0$. $D = b^2 - 4ac = (\cos p)^2 - 4

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$p \in (0,\pi )$

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(C) Given equation: $(\cos p - 1){x^2} + (\cos p)x + \sin p = 0$. Since the roots are real,the discriminant $D \ge 0$. $D = b^2 - 4ac = (\cos p)^2 - 4(\cos p - 1)(\sin p) \ge 0$. $\cos^2 p - 4\sin p \cos p + 4\sin p \ge 0$. For the quadratic equation to be valid,the coefficient of $x^2$ must not be zero,so $\cos p - 1 
eq 0$,which implies $p 
eq 2n\pi$. Analyzing the inequality,for $p \in (0, \pi)$,$\sin p > 0$ and $(1 - \sin p) \ge 0$. Thus,the condition $D \ge 0$ is satisfied for $p \in (0, \pi)$.

If the roots of the given equation $(\cos p - 1){x^2} + (\cos p)x + \sin p = 0$ are real,then

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