The number of real roots of the equation $x^2+x+3+2\sin x = 0, x ∈[-π, π]$, is ______.

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We have, $x^2+x+3+2 \sin x = 0$ $⇒x^2+x+3=-2\sin x ⇒ (x+\frac{1}{2})^2 +\frac{11}{4}=-2\sin x$ We observe that $LHS >\frac{11}{4}>2$ for all x whereas RHS lies between-2 and 2.Hence, the given equation has no solution.