Question

If $0\le \theta \le 2\pi$ then solve the equation $(\sin 2\theta +\sqrt{3}\cos 2\theta {)}^2-5=\cos (2\theta -\frac{\pi }{6})$.

Answer 1

Ist term $=\left[2\cos \right(2\theta -\pi /6){]}^2$
$4{\cos }^2(2\theta -\frac{\pi }{6})-5=\cos (2\theta -\frac{\pi }{6})$
or $(4c-5)(c+1)=0$
$\therefore c=\cos (2\theta -\frac{\pi }{6})=-1$ as $c\ne \frac{5}{4}$ i.e., $>1$
$\therefore 2\theta -\frac{\pi }{6}=\pi$, $3\pi$ only as $0<2\theta <4\pi$
or $2\theta =\frac{7\pi }{6},\frac{19\pi }{6}$
$\therefore \theta =\frac{7\pi }{12},\frac{19\pi }{12}$