Find the number of roots of the equation, x3+x2+2x+sin⁡x=0 in [−2π,2π]

Question

Joy Compton · Accepted Answer

∀x(f′(x)=3x2+2x+2+cos⁡x&amp;gt;0) (the polynomial term is always at least 53), so note that f(−2π)&amp;lt;0,  f(2π)&amp;gt;0 and conclude that there is exactly 1 root.

Gwendolyn Meyer · Accepted Answer

For x&amp;lt;0, sin⁡x=x−x33!+O(x5)&amp;gt;x:   x3+x2+2x−sin⁡x&amp;lt;x3+x2+2x−x=x(x2+x+1)&amp;lt;0,   for x&amp;gt;0,sin⁡x&amp;lt;x:   x3+x2+2x−sin⁡x&amp;gt;x3+x2+2x−x=x3+x2+x&amp;gt;0

Find the number of roots of the equation, x3+x2+2x+sin⁡x=0 in [−2π,2π]

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