The equation $(c o s p - 1) x^{2} + (c o s p) x + s i n p = 0$ in variable $x$ has real roots, if $p$ belongs to the interval

$(0, π)$

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$(- \frac{π}{2}, \frac{π}{2})$

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$(- π, 0)$

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$(0, 2 π)$

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Solution

The correct option is A
$(0, π)$

$(c o s p - 1) x^{2} + (c o s p) x + s i n p = 0 . . . . . (1)$

Discriminant of $(1)$ is given by $D = c o s^{2} p - 4 (c o s p - 1) s i n p$ $= c o s^{2} p + 4 (1 - c o s p) s i n p$

Note that $c o s^{2} p \geq 0, 1 - c o s p \geq 0.$
Thus $D \geq 0$ if $sin p \geq 0 \Rightarrow p \in (0, π) .$

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