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Discriminant

The equation ...

Question

The equation $(c o s p - 1) x^{2} + c o s p x + s i n p = 0$ in x has real roots. Then the set of values of p is

[0, 2 π]

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[- π, 0]

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[- \frac{π}{2}, \frac{π}{2}]

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[0, π]

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Solution

The correct option is D $[0, π]$
$(c o s p - 1) x^{2} + c o s p . x + s i n p = 0$
$D = c o s^{2} p - 4 s i n p (c o s p - 1) \geq 0$
if sin p must be positive.
$0 \leq p \leq π$

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