calculator: https://www.desmos.com/calculator solve 3 sin(2x) = 2 for the two smallest positive solutions…

Explanation:

Step1: Isolate the sine - function

Divide both sides of the equation $3\sin(2x)=2$ by 3. We get $\sin(2x)=\frac{2}{3}$.

Step2: Use the inverse - sine function

If $\sin(2x)=\frac{2}{3}$, then $2x = \sin^{- 1}(\frac{2}{3})+2k\pi$ or $2x=\pi-\sin^{- 1}(\frac{2}{3}) + 2k\pi$, where $k\in\mathbb{Z}$. First, find $\sin^{- 1}(\frac{2}{3})\approx0.7297$ (in radians). For $2x=\sin^{- 1}(\frac{2}{3})+2k\pi$: When $k = 0$, $2x=\sin^{- 1}(\frac{2}{3})\approx0.7297$, then $x=\frac{\sin^{- 1}(\frac{2}{3})}{2}\approx0.365$. For $2x=\pi-\sin^{- 1}(\frac{2}{3})+2k\pi$: When $k = 0$, $2x=\pi - 0.7297\approx3.1416 - 0.7297=2.4119$, then $x=\frac{\pi-\sin^{- 1}(\frac{2}{3})}{2}\approx1.206$.

Answer:

$x\approx0.365,x\approx1.206$