# Question #91a43

Jun 5, 2016

A rounded 5-sd set of solutions is
$x = \left(4 n + 0.76528\right) \pi , n = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$.
Of course, $x = 0 , \pm 2 \pi , \pm 4 \pi , \pm 6 \pi , . .$are also solutions.

#### Explanation:

Let $y \left(x\right) = \sin x + 1.5 \sin \left(1.5 x\right)$

This represents the compounded oscillation of the separate

oscillations for sin x and 1.5 sin (1.5 x ) of periods $2 \pi \mathmr{and} \frac{4 \pi}{3}$,

respectively.

Now, $y \left(x + 4 \pi\right) = y \left(x\right)$. So, y is periodic with period $4 \pi$.

I have obtained one approximate solution from

$y \left(137 , {753}^{o}\right) = y \left(0.76528 \pi\right) = 0.000001$, nearly.

This 5-sd solution $\in \left(0 , 2 \pi\right)$ was obtained by using a

convergent root-bracketing (optimized bisection) method.

Correspondingly, the set of genera(.solutions is obtained as

$x = \left(4 n + 0.76528\right) \pi , n = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

Likewise, another such root $\in \left(2 \pi , 4 \pi\right)$ could be approximated.

Correspondingly, one more set of general solutions

could be obtained.

Indeed, y becomes 0, for x= $2 n \pi , n = 0 , \pm 1 , \pm 2 , \pm 3 , . .$