Question

Solve: $2(sin\theta -\cos 2\theta )-sin2\theta (1+2sin\theta )+2cos\theta =0$

Answer 1

$2(\sin \theta -\cos 2\theta )-\sin 2\theta (1+2\sin \theta )+2\cos \theta =0$

$\Rightarrow 2\sin \theta -2\cos \theta -\sin 2\theta -2\sin 2\theta \sin \theta +2\cos \theta =0$

$\Rightarrow 2\sin \theta (1-\sin 2\theta )-2\cos 2\theta -\sin 2\theta +2\cos \theta =0$

$\Rightarrow 2\sin \theta -4{\sin }^2\theta \cos \theta -2\cos 2\theta -2\sin \theta \cos \theta +2\cos \theta =0$

$\Rightarrow 2\sin \theta -4{\sin }^2\theta \cos \theta -2{\cos }^2\theta +2{\sin }^2\theta -2\sin \theta \cos \theta +2\cos \theta =0$
$\Rightarrow 2\sin \theta +2\cos \theta [1-2{\sin }^2\theta ]-2\cos 2\theta -2\sin \theta \cos \theta =0$

$\Rightarrow 2\sin \theta [1-\cos \theta ]+2\cos \theta .2\cos 2\theta -2\cos 2\theta =0$

$\Rightarrow 2\sin \theta [1-\cos \theta ]+2\cos 2\theta [2\cos \theta -1]=0$

$=2\sin \theta [1-\cos \theta ]+2\cos 2\theta [\cos \theta -1]+2\cos 2\theta \cos \theta =0$

$=(\cos \theta -1)(2\cos 2\theta -2\sin \theta )+2\cos 2\theta \cos \theta =0$

$\Rightarrow \cos 2\theta =\sin \theta \Rightarrow \theta =\frac{\pi }{6}$