Question

Find $\theta$:

$3-2\cos \theta -4\sin \theta -\cos 2\theta +\sin 2\theta =0$

Answer 1

$3-2\cos \theta -4\sin \theta -(1-2{\sin }^2\theta )+2\sin \theta \cos \theta =0$
$2{\sin }^2\theta -4\sin \theta +2-2\cos \theta +2\sin \theta \cos \theta =0$
$({\sin }^2\theta -2\sin \theta +1)+\cos \theta (\sin \theta -1)=0$
$(\sin \theta -1{)}^2+\cos \theta (\sin \theta -1)=0$
$\therefore (\sin \theta -1)(\sin \theta -1)+\cos \theta )=0$
$\sin \theta =1=\sin \frac{\pi }{2}$
$\therefore \theta =n\pi +(-1{)}^n\frac{\pi }{2}$
$\cos \theta +\sin \theta =1$
Divide by $\sqrt{1+1}=\sqrt{2}$
$\frac{1}{\sqrt{\left(2\right)}}\cos \theta +\frac{1}{\sqrt{\left(2\right)}}\sin \theta =\frac{1}{\sqrt{\left(2\right)}}$
or $\cos \theta \cos \frac{\pi }{4}+\sin \theta \sin \frac{\pi }{4}=\frac{1}{\sqrt{\left(2\right)}}$
or $\cos (\theta -\frac{\pi }{4})=\cos \frac{\pi }{4}$
$\therefore \theta -\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4}$
$\therefore \theta =2n\pi$ or $2n\pi +\frac{\pi }{4}+\frac{\pi }{4}=2n\pi \pm \frac{\pi }{2}$
We could also write it as
$\sin (\theta +\frac{\pi }{4})=\frac{1}{\sqrt{\left(2\right)}}=\sin \frac{\pi }{4}$
$\therefore \theta +\frac{\pi }{4}=n\pi +(-1{)}^n\frac{\pi }{4}$
$\therefore =n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{4}$.