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Solve $\cos 2x - 3\sin x - 1 = 0, \quad 0^{\circ} \le x \le 360^{\circ}$

\begin{align} \cos 2x - 3\sin x - 1 = 0 &\iff 1 - 2\sin^2 x - 3\sin x - 1 = 0 \\ &\iff- 2\sin^2 x - 3\sin x = 0 \\ &\iff2\sin^2 x + 3\sin x =0\\ &\iff\sin x(2\sin x + 3) =0 \\ &\iff\sin x = 0 \lor 2 \sin x + 3= 0 \end{align} I could go on but the book gives the answer, $0^{\circ}, 180^{\circ}, 360^{\circ}$ and I am mystified as to where these answers have come from.

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    $\begingroup$ There is no (real) solution to $2\sin x+3=0$ since $\sin x$ is always between $-1$ and $1$. So $\sin x=0$, and you know the angles whose sine is $0$. $\endgroup$ – André Nicolas Apr 14 '16 at 6:07
  • $\begingroup$ I do not understand why 360 is here for $\sin x = 0$, sin is negative in 4th quadrant. $\endgroup$ – dagda1 Apr 14 '16 at 6:18
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    $\begingroup$ @dadga1 $\sin 360^{\circ} = 0$. Remember, it is a full rotation $\endgroup$ – MathematicsStudent1122 Apr 14 '16 at 6:24
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$\sin x =0 \Leftrightarrow x =\pi n (0^{\circ}, 180^{\circ}, 360^{\circ})$

$2\sin x=-3 \Rightarrow \sin x = -\frac 32 - $impossible

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