Explanation

(b)
Equation first can be written as
$x\sin{a + y\ \times}\ 2\sin{a\cos{a + \mathcal{z} \times \sin{a\ \left( 3 - 4\sin^{2}a \right)}}} = 2\ \times \ 2\sin{a\cos{a\cos{2a}}}$
$\Rightarrow x + 2y\cos{a + \mathcal{z}\left( 3 + 4\cos^{2}{a - 4} \right)} = 4\cos{a\ \left( 2\cos^{2}{a - 1} \right)\ as\sin{a\ \neq 0}}$
$\Rightarrow 8\cos^{3}{a - 4\mathcal{z}\cos^{2}{a - (2y + 4)\cos{a + \left( \mathcal{z -}x \right) = 0}}}$
$\Rightarrow \cos^{3}{a - \left( \frac{\mathcal{z}}{2} \right)\cos^{2}{a - \left( \frac{y + 2}{4} \right)\cos{a + \left( \frac{\mathcal{z -}x}{8} \right) = 0}}}$
Which shows that $\cos{a\ }$is a root of the equation
$t^{3} - \left( \frac{\mathcal{z}}{2} \right)t^{2} - \left( \frac{y + 2}{4} \right)t + \left( \frac{\mathcal{z -}x}{8} \right) = 0$
Similarly, from second and third equation we can verify that $\cos{b\ }$and $\cos{c\ }$are the roots of the given equation