$\sin x + \sin 2x + \sin 3x = 2(\sin 2x)(\cos x) + \sin 2x = (\sin 2x)(2\cos x + 1)$
$\sin 2x = 0\ \vee\ 2\cos x + 1 = 0\ \Rightarrow\ x = \dfrac{2\pi}{3}$
$S = \left\{\dfrac{2\pi}{3}\right\}$
Organização sem fins lucrativos, voltada para a pesquisa e educação em Matemática.
$\sin x + \sin 2x + \sin 3x = 2(\sin 2x)(\cos x) + \sin 2x = (\sin 2x)(2\cos x + 1)$
$\sin 2x = 0\ \vee\ 2\cos x + 1 = 0\ \Rightarrow\ x = \dfrac{2\pi}{3}$
$S = \left\{\dfrac{2\pi}{3}\right\}$
$2\sin \theta = 3\tan^2 \theta$
$\dfrac{2}{3} = \dfrac{\sin \theta}{\cos^2 \theta} = \dfrac{\sin \theta}{1 - \sin^2 \theta}$
$2 - 2\sin^2 \theta = 3\sin \theta$
$2\sin^2 \theta + 3\sin \theta - 2 = 0$
$\sin \theta = \dfrac{1}{2}\ \Rightarrow\ \fbox{$\cos \theta = \dfrac{\sqrt{3}}{2}$}$