Solving $\frac{x^{3} - 2x^{2} + x + 4}{x+1} = \frac{7}{4}\sin\frac{\pi x}{3}$

Question

I'd like to simply solve this equation: $$\frac{x^{3} - 2x^{2} + x + 4}{x+1} = \frac{7}{4}\sin\frac{\pi x}{3}$$ What makes it difficult for me is the conversion between the polynomial expression for $x$ on the left side that should be equal to a trigonometric function of a product of $x$ and a radian value. I've heard of Taylor and Maclaurin's methods, but I don't think it would help. It's probably something quite basic, but it's just new for me. Thanks a lot in advance!

Answer 1

Hint: prove that $$x=\frac{3}{2}$$ is the only real solution. Solve the inequality $$\left|\frac{x^3-2x^2+x+4}{x+1}\right|\le \frac{7}{4}$$

Answer 2

Hint: $\;x^3-2x^2+x+4\,$ has the root $\,x=-1\,$ (easy to find by inspection, or the rational root theorem), then the numerator factors as $\,x^3-2x^2+x+4=(x+1)(x^2 - 3 x + 4)\,$, and the fraction on the LHS simplifies to $\,\dfrac{x^3-2x^2+x+4}{x+1}=x^2-3x+4\,$. All that's left is to check the range of the quadratic.

Answer 3

Okay, what I've come up with was the following idea of working it all out: As already mentioned, $$\frac{x^3 - 2x^2 + x + 4}{x+1} = x^2 - 3x + 4$$ which, when factorized right, gives $$x^2 - 3x + 4 = (x - \frac{3}{2})^2 + \frac{7}{4} \ge \frac{7}{4}$$ whereas the right side $$\frac{7}{4}sin\frac{\pi x}{3} \le \frac{7}{4}$$

Which inevitably leads to the conclusion that equality is only possible when $$\begin{array}{|l} (x - \frac{3}{2})^2 + \frac{7}{4} = \frac{7}{4} \\ \frac{7}{4}sin\frac{\pi x}{3} = \frac{7}{4} \end{array}$$ an only solution to which is $x = \frac{3}{2}$. Thanks for the help!