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



Answered question


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

Answer & Explanation

Joy Compton

Joy Compton

Beginner2022-01-25Added 13 answers

x(f(x)=3x2+2x+2+cosx>0) (the polynomial term is always at least 53), so note that f(2π)<0,  f(2π)>0 and conclude that there is exactly 1 root.
Gwendolyn Meyer

Gwendolyn Meyer

Beginner2022-01-26Added 11 answers

For x<0, sinx=xx33!+O(x5)>x:
for x>0,sinx<x:

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