The-number-of-solution-s-of-the-equation-x-3-x-2-4x-2sinx-0-in-0-2pi-is-are-

Question Number 23231 by Tinkutara last updated on 27/Oct/17

The number of solution (s) of the equation

x^{3} + x^{2} + 4 x + 2 \sin x = 0 in [0, 2 π], is / are

Answered by ajfour last updated on 28/Oct/17

O n e .

x = 0 i s o b v i o u s l y a s o l u t i o n, a n d

b e y o n d t h a t \dots (b u t f i r s t l e t)

f (x) = x^{3} + x^{2} + 4 x + 2 \sin x

f^{'} (x) = 3 x^{2} + 2 x + 4 + 2 \cos x

= 3 {(x + \frac{1}{3})}^{2} - \frac{1}{3} + 2 + 2 (1 + \cos x)

= 3 {(x + \frac{1}{3})}^{2} - \frac{1}{3} + 2 + 4 \cos^{2} (\frac{x}{2})

= 3 {(x + \frac{1}{3})}^{2} + \frac{5}{3} + 4 \cos^{2} (\frac{x}{2}) > 0

s o n o m o r e s o l u t i o n s .

Commented by Tinkutara last updated on 28/Oct/17

Thank you very much Sir!