Solve $\cos 2x - \sin x = 0$.
\begin{align}\cos 2x = 1 - 2\sin^2 x &\iff 1 -2 \sin^2 x - \sin x = 0 \\ &\iff -2 \sin^2 x - \sin x + 1 = 0 \\ &\iff 2 \sin^2 x + \sin x - 1 = 0 \\ &\iff \sin x(2\sin x + 1) -1 =0\\ &\iff (2\sin x + 1)(\sin - 1) =0\\ &\iff \sin x = -\frac{1}2 \lor \sin x = 1 \end{align}
I could go on but the book gives the answer to this question $x = 30, 150$ and $270$ so I have obviously gone wrong somewhere.
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$\begingroup$You are fine up to here:
$$ 2 \sin^2 x+\sin x-1 = 0 $$
You then write this as
$$ (\sin x)(2 \sin x+1)-1 = 0 $$
which is true (though not quite helpful), but then you somehow factor this as
$$ (2\sin x+1)(\sin x-1) = 0 $$
which is not right. You should instead rewrite $2 \sin^2 x + \sin x - 1 = 0$ as
$$ (2 \sin x - 1)(\sin x + 1) = 0 $$
which will yield the answers in the book.
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