The number of solutions of the equation $x^{3} + 2 x^{2} + 5 x + 2 c o s x = 0$ in $[0, 2 π]$ is

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Solution

Solution : $x^{3} + 2 x^{2} + 5 x + 2 c o s x = 0$
$x^{3} + 2 x^{2} + 5 x = - 2 c o s x$
$y = x (x^{2} + 2 x + 5) = - 2 c o s x$
$y = x (x^{2} + 2 x + 5)$
if $x \in [0, 2 π]$ then number of solution is zero.

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