4

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?

grg
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Irtiza
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1 Answers1

1

$2\pi+\sin^{-1}\left(\frac{3}{4}\right)$ 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

$$x=\frac{2\sin\theta}{\sqrt{3}}$$

KittyL
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