Find the inflection points of the curve y = sin³x

Find the inflection points of the curve $y = \sin^{3}x$.

Step 1: The second derivative

\[ y' = 3\sin^{2}x\cos x, \]

\[ y'' = 3\left(2\sin x\cos^{2}x-\sin^{3}x\right) = 3\sin x\bigl(2\cos^{2}x-\sin^{2}x\bigr) = 3\sin x\bigl(2-3\sin^{2}x\bigr). \tag{1} \]

Step 2: Candidates

\[ \sin x = 0 \;\Longrightarrow\; x = n\pi; \qquad \sin^{2}x = \frac23 \;\Longrightarrow\; \sin x = \pm\sqrt{\frac23}. \]

Step 3: All of them are inflections

At $x = n\pi$ the factor $\sin x$ changes sign while $2-3\sin^{2}x\to2\gt 0$ stays positive nearby, so $y''$ changes sign. At the points where $\sin^{2}x = \tfrac23$ the factor $\sin x$ is non-zero and keeps its sign, while $2-3\sin^{2}x$ crosses zero transversally. In both cases the concavity reverses.

Step 4: The three families of points

With $\beta = \arcsin\sqrt{\tfrac23} = 0.9553166$, $\sin x = 0:\quad \bigl(n\pi,\;0\bigr)$.

\[ \sin x = +\sqrt{\tfrac23}:\quad \left(\beta+2k\pi \ \text{ or } \ \pi-\beta+2k\pi,\ \ \frac{2\sqrt6}{9}\right); \]

\[ \sin x = -\sqrt{\tfrac23}:\quad \left(-\beta+2k\pi \ \text{ or } \ \pi+\beta+2k\pi,\ \ -\frac{2\sqrt6}{9}\right), \] since $\left(\sqrt{\tfrac23}\right)^{3} = \left(\tfrac23\right)^{3/2} = \dfrac{2\sqrt2}{3\sqrt3} = \dfrac{2\sqrt6}{9}$.

Step 5: Verification

\[y''(0.5) = 3\cdot0.4794\bigl(2-3\cdot0.2298\bigr) = 1.885\gt 0; \qquad y''(1.5) = 3\cdot0.9975\bigl(2-2.9850\bigr) = -2.948\lt 0 \;\;\checkmark\]

\[\beta = 0.9553166 \text{ lies between them, and } \sin^{2}\beta = \tfrac23 \;\;\checkmark\]

\[\text{At } x = n\pi \text{ also } y' = 0:\ \text{these are the stationary inflections of Problem 431} \;\;\checkmark\]