How can i solve $x\sin^2(x) + 5x\sin(x) + 4 = 0 $?

Question

If I have an equation such as $$ x\sin^2(x) + 5x\sin(x) + 4 = 0 $$ with trigonometric functions, how can I solve it?

Answer 1

Solving the quadratic equation we have

$$ 2x\sin(x)+ 5x\mp \sqrt{25x^2-16x}=0$$

There are two separate transcendental equations to be solved numerically.

Answer 2

Your equation is a quadratic in $\sin(x)$ with coefficients $x$, $5x$, and $4$. We can use the quadratic formula:

\begin{align} \sin(x) \, &= \, \displaystyle\frac{-5x \pm \sqrt{25x^2-16x}}{2x}\\\\ \sin(x) \, &= \, \displaystyle\frac{-5x \pm \sqrt{x^2\left(25-\displaystyle\frac{16}{x}\right)}}{2x}\\\\ \sin(x) \, &= \, \displaystyle\frac{-5 \pm \sqrt{25-\displaystyle\frac{16}{x}}}{2}\\ \end{align}

Note that the negative case of the right-hand-side expression will never exceed $-5/2$, and so cannot equal an output of $\sin(x)$. So solutions must satisfy $$\sin(x) \, = \, \displaystyle\frac{-5 + \sqrt{25-\displaystyle\frac{16}{x}}}{2}$$

This doesn't lead to a closed form solution, but graphing the left-hand-side and the right-hand-side shows us the solution set and that it is infinite:

If you're interested in a specific solution for some application, you may find a nearby point via this graph, then use numerical methods to efficiently obtain arbitrary precision.