<s>[INST] Solve the equation $x^3-x+\frac16\sqrt3=0$ 
8 (i) By first expanding $\sin(2\theta+\theta)$, show that $$\sin3\theta=3\sin\theta-4\sin^3\theta\tag4$$
  (ii) Show that, after making the substitution $x=\frac{2\sin\theta}{\sqrt3}$, the equation $x^3-x+\frac16\sqrt3=0$
           can be written in the form $\sin3\theta=\frac34$. $\tag1$
 (iii) Hence solve the equation $$x^3-x+\frac16\sqrt3=0$$
          giving your answers correct to 3 significant figures. $\tag4$

This equation in part 3. I can solve this by using
$$\begin{align}
\sin(3\theta)&=\left(\frac{3}{4}\right)\\
3\theta&=\sin^{-1}\left(\frac{3}{4}\right)\\
3\theta&=\pi-\sin^{-1}\left(\frac{3}{4}\right)\\
3\theta&=2\pi+\sin^{-1}\left(\frac{3}{4}\right)
\end{align}$$
…and then getting values for $\theta$ after dividing these expressions by $3$, but the marking scheme says something else:

(iii) Carry out a correct method to find a value of $x$ $\qquad\mathrm M1$
          Obtain answers $0.322, 0.799, -1.12$ $\qquad\mathrm A1+\mathrm A1+\mathrm A1\quad[4]$
          [Solutions with more than 3 answers can only earn a maximum of $\mathrm A1+\mathrm A1$.]

What am I missing?
 [/INST] $2\pi+\sin^{-1}\left(\frac{3}{4}\right)$ gives you the same $\sin$ value as $\pi-\sin^{-1}\left(\frac{3}{4}\right)$. 
You can do this: 
$$3\theta = 2k\pi+\frac{\pi}{6} \text{ or } 2k\pi+\frac{5}{6}\pi$$
$$\theta = \frac{2}{3}k\pi+\frac{\pi}{18} \text{ or } \frac{2}{3}k\pi+\frac{5}{18}\pi$$
Let $k=0,1,2$. Although you get $6$ values, there are only $3$ that have distinct $\sin$ values.
Now compute 
$$x=\frac{2\sin\theta}{\sqrt{3}}$$
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