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

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{solution}\left(\mathrm{s}\right)\mathrm{of}\mathrm{the}\mathrm{equation}$$$$x^3+x^2+4x+2\sin x=0\mathrm{in}[0,2\pi ],\mathrm{is}/\mathrm{are}$$

Answered by ajfour last updated on 28/Oct/17

$$One.$$$$x=0isobviouslyasolution,and$$$$beyondthat...\left(butfirstlet\right)$$$$f\left(x\right)=x^3+x^2+4x+2\sin x$$$$f{}^{\prime }\left(x\right)=3x^2+2x+4+2\cos x$$$$=3{(x+\frac{1}{3})}^2-\frac{1}{3}+2+2(1+\cos x)$$$$=3{(x+\frac{1}{3})}^2-\frac{1}{3}+2+4\cos {}^2\left(\frac{x}{2}\right)$$$$=3{(x+\frac{1}{3})}^2+\frac{5}{3}+4\cos {}^2\left(\frac{x}{2}\right)>0$$$$sonomoresolutions.$$

Commented by Tinkutara last updated on 28/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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