Edexcel IAL Mathematics FP2.3.2 De Moivre's theorem

$$(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$$
B1
$$\begin{aligned} &=\cos^5\theta+5\cos^4\theta(i\sin\theta)+10\cos^3\theta(i\sin\theta)^2\\ &\quad+10\cos^2\theta(i\sin\theta)^3+5\cos\theta(i\sin\theta)^4+(i\sin\theta)^5 \end{aligned}$$
M1
$$=\cos^5\theta+5i\sin\theta\cos^4\theta-10\sin^2\theta\cos^3\theta-10i\sin^3\theta\cos^2\theta+5\sin^4\theta\cos\theta+i\sin^5\theta$$
A1
Equating imaginary parts,
$$\sin5\theta=5\sin\theta\cos^4\theta-10\sin^3\theta\cos^2\theta+\sin^5\theta$$$$=5\sin\theta(1-\sin^2\theta)^2-10\sin^3\theta(1-\sin^2\theta)+\sin^5\theta$$
M1
$$\sin5\theta=16\sin^5\theta-20\sin^3\theta+5\sin\theta$$
A1*

Let $x=\sin \theta \quad 16 x^{5}-20 x^{3}+5 x=-\frac{1}{5} \Rightarrow \sin 5 \theta=\ldots$

M1A1

$\Rightarrow \theta=\frac{1}{5} \sin ^{-1}\left( \pm \frac{1}{5}\right)=38.306$ (or -2.307,69.692.110.306,141.693,182.306 )
(or in radians $-0.0402 \ldots 0.6685 \ldots, 1.216 \ldots, 1.925 \ldots, 2.473 \ldots$ )

dM1

Two of (awrt) $x=\sin \theta=-0.963,-0.555,-0.040,0.620,0.938$

A1

All of (awrt) $x=\sin \theta=-0.963,-0.555,-0.040,0.620,0.938$

A1 (5)