All Obtained Different Results? Who's Right?

$2\sin x + \cos x = 1$

$\implies \dfrac{2}{\sqrt{5}}\sin x + \dfrac{1}{\sqrt{5}}\cos x = \dfrac{1}{\sqrt{5}}$

Writing $\dfrac{2}{\sqrt{5}} \text{ as }\cos \alpha \implies \sin \alpha = \dfrac{1}{\sqrt{5}}$

$\therefore \cos \alpha \sin x + \sin \alpha \cos x = \sin \alpha$

$\implies \sin(x + \alpha) = \sin \alpha$

$\implies x + \alpha = n\pi + (-1)^n \alpha$

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Case 1:

If $n = 2k, k \in \mathbb{Z}:$

$x = 2k\pi$

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Case 2:

If $n = 2k+1, k \in \mathbb{Z}:$

$x = 2k\pi + \pi - 2\alpha$

Now, we have to verify the alternate forms of $\pi - 2\alpha$ according to the choices given.

We can directly see that $\boxed{\text{Antony is correct}}$ as $\sin \alpha = \dfrac{1}{\sqrt{5}} \iff \alpha = \arcsin \dfrac{\sqrt{5}}{5}$

We also know that:

$\tan \alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{1}{2} \implies \cot \alpha = 2$

$\implies \alpha = \text{arccot } 2 = \dfrac{\pi}{2} - \arctan 2 \implies \pi - 2\alpha = 2\arctan 2$

$\therefore \boxed{\text{Barbara is also correct}}.$

$\sin 2\alpha = 2\sin\alpha\cos\alpha = 2\cdot\dfrac{2}{\sqrt{5}}\cdot\dfrac{1}{\sqrt{5}} = \dfrac{4}{5}$

$\implies 2\alpha = \arcsin\dfrac{4}{5} \iff \pi - 2\alpha =\pi - \arcsin\dfrac{4}{5}$

$\therefore \boxed{\text{Maeve is also correct}}.$

Also,

$\pi - \arcsin\dfrac{4}{5} = \beta \text{(let)}$

(As $\arcsin(x) \in [0, \dfrac{\pi}{2}] for x > 0, \beta \text{ must be an obtuse angle.}$ )

$\sin\Bigg(\pi - \arcsin\dfrac{4}{5}\Bigg) = \sin\beta$

$\implies \sin\beta = \dfrac{4}{5}$

$\implies \cos\beta = -\dfrac{3}{5} \hspace{3 cm}(\because \beta \text{ is an obtuse angle, its cosine is negative.)}$

$\implies \beta = \arccos \Bigg(-\dfrac{3}{5}\Bigg)$

$\therefore \boxed{\text{Michaelle is also correct}}.$