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(C) Given the quadratic equation $(\cos p - 1)x^2 + (\cos p)x + \sin p = 0$.For the roots to be real,the discriminant $D$ must be greater than or equa

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$p \in (0, \pi)$

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(C) Given the quadratic equation $(\cos p - 1)x^2 + (\cos p)x + \sin p = 0$.For the roots to be real,the discriminant $D$ must be greater than or equal to zero $(D \ge 0)$.$D = b^2 - 4ac = (\cos p)^2 - 4(\cos p - 1)(\sin p) \ge 0$.Expanding this,we get $\cos^2 p - 4\sin p \cos p + 4\sin p \ge 0$.This can be rewritten as $(\cos p - 2\sin p)^2 - 4\sin^2 p + 4\sin p \ge 0$.$(\cos p - 2\sin p)^2 + 4\sin p(1 - \sin p) \ge 0$.Since $(\cos p - 2\sin p)^2 \ge 0$ for all real $p$,the inequality holds if $4\sin p(1 - \sin p) \ge 0$.Since $(1 - \sin p) \ge 0$ for all real $p$,we require $\sin p \ge 0$.In the interval $(0, 2\pi)$,$\sin p \ge 0$ holds 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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