So I have $\sin^3x = \frac 34 \sin x$.
Can you expand so the answer is either $\sin x(\sin^2x +\frac 34)$ which leads to the answer $\frac 12 + 2n\pi$ or that $\sin^3x = \frac 14(3\sin x-\sin^3x) - \frac 34\sin x$ which leads to the answer $0 + 2n \pi$.
Is that correct by any chance?
$$\sin^3 x = \frac34 \sin x$$
$$\sin x(\sin^2 x - \frac 34) = 0$$
Hence $\sin x = 0$ or $\sin^2 x - \frac 34=0$
First possibility gives $x = n\pi, n \in \mathbb{Z}$
Second possibility gives $\sin x = \pm \frac {\sqrt 3}{2}$. Can you solve this final step? Hint: think of the special 30-60-90 right triangle, and consider all quadrants.
$t = \sin x \to t^3-\dfrac{3t}{4} = 0\to 4t^3-3t=0\to t(4t^2-3)=0\to t = 0,\pm \dfrac{\sqrt{3}}{2}\to x = n\pi, \pm \dfrac{\pi}{3}+2n\pi,\pm\dfrac{2\pi}{3}+2n\pi, n \in \mathbb{Z}$
I would go with your first expression (but you forgot the $=0$ part), because
$$\sin^3x = \frac 34 \sin x \iff \sin x\ \left(\sin^2x -\frac 34\right)=0$$
The equation on the right implies that either $\sin x = 0$ or $\sin x=\pm \frac{\sqrt{3}}{2}$, and from here you can use the unit circle to solve for $x$.
Move 3/4sinx to the left side and regroup to get
sinx((sin^2)(x) - 3/4) = 0
So all values of x such that sinx = 0 or sinx = sqrt(3)/2 satisfy this equality.
These values are all n*pi, pi/3(1+6n), pi/3(5+6n) where n=(0,1,2,...)
$$sin^3x=\frac{3}{4}sinx \\$$multiply by $4$ $$4sin^3x=3sinx \\3sinx -4sin^3x=0\\sin(3x)=0\\3x=k\pi\\x=\frac{\pi}{3}$$
Notice, we have $$\sin^3x = \frac 34 \sin x$$ $$\sin^3x -\frac 34 \sin x=0$$ $$\sin x\left(\sin^2 x-\frac{3}{4}\right)=0$$ $$\text{if}\ \sin x=0\ \implies \color{blue}{x=n\pi}$$ $$\text{if}\ \sin^2 x-\frac{3}{4}\iff \sin^2x=\left(\frac{\sqrt{3}}{2}\right)^2 \iff \sin^2x=\left(\sin \frac{\pi}{3}\right)^2\ \implies \color{blue}{x=n\pi\pm \frac{\pi}{3}}$$ Where, $n$ is an integer.
It turns out that $\sin(3x)=3\sin(x)-4\sin^3(x)$ so that $3x=k\pi$.