Find the number of solutions of the equation, $s i n = x^{2} + x + 1$

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Solution

Let: $f (x) = s i n x a n d g (x) = x^{2} + x + 1 = (x + \frac{1}{2})^{2} + \frac{3}{4}$

Discriminant of $x^{2} + x + 1$

$D = b^{2} - 4 a c = 1 - 4 = - 3$ (No real solution)

since discriminant is negative $x^{2} + x + 1$ will have no real solution.

vertex of graph $x^{2} + x + 1$ is

X - coordinate $= \frac{- b}{2 a} = \frac{- 1}{2}$

Y - coordinate $= \frac{- D}{4 a} = \frac{- (- 3)}{4 \times 1} = \frac{3}{4}$

and at $x = 0, y = 0 + 0 + 1 = 1$

at $x = 1, y = 1 + 1 + 1 = 3$

Graph could be shown as;

which do not intersect at any point,therefore No solution.

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