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