Question

$\sqrt{3}\cos \theta +\sin \theta =1$ for $-2\pi <\theta <2\pi$.

Solve for $\theta$

Answer 1

As above, $\cos (\theta -\frac{\pi }{6})=\frac{1}{2}=\cos \frac{\pi }{3}$
$\therefore \theta -\frac{\pi }{6}=2n\pi \pm \frac{\pi }{3}$
$\theta =2n\pi +\frac{\pi }{6}+\frac{\pi }{3}$ or $2n\pi +\frac{\pi }{6}-\frac{\pi }{3}$
$=2n\pi +\frac{\pi }{2}$ or $2n\pi -\frac{\pi }{6}$
$\theta =(4n+1)\frac{\pi }{2}$ or $(12n-1)\frac{\pi }{6}$
Numerators are in A.P. of $d=4$ and $d=12$ $-2\pi <\theta <2\pi$
$\therefore \theta =\frac{(-7,-3,1,5,....)}{4}\cdot 2\pi$
and $\theta =\frac{(-25,-13,-1,11,23,25,...)}{12}\cdot 2\pi$
$\therefore \theta =(-\frac{3}{4},\frac{1}{4})2\pi ,(-\frac{1}{12},\frac{11}{12})\cdot 2\pi$
$\therefore \theta =\frac{-3\pi }{2},\frac{-\pi }{6},\frac{\pi }{2}$ and $\frac{11\pi }{6}$.