Question

If $2\sin (\theta +\frac{\pi }{3})=\cos (\theta -\frac{\pi }{3})$, and $\tan \theta +\sqrt{3}=0$. Find the value of $\theta$.

Answer 1

$2sin(\theta +\frac{\pi }{3})=cos(\theta -\frac{\pi }{3})$

$2(sin\theta cos\frac{\pi }{3}+cos\theta sin\frac{\pi }{3})=cos\theta cos\frac{\pi }{3}+sin\theta sin\frac{\pi }{3}$

$2(\frac{1}{2}\theta +\frac{\sqrt{3}}{2}cos\theta )=\frac{1}{2}cos\theta +\frac{\sqrt{3}}{2}sin\theta$

$\frac{1}{2}[2sin\theta +2\sqrt{3}cos\theta -cos\theta -\sqrt{3}sin\theta ]=0$

$\left[\right(2-\sqrt{3})sin\theta +(2\sqrt{3}-1\left)cos\theta \right]$

$tan\theta =\frac{sin\theta }{cos\theta }=\frac{1-2\sqrt{3}}{2-\sqrt{3}}$

putting the value of $tan\theta$ in

$tan\theta +\sqrt{3}$

$\Rightarrow \frac{1-2\sqrt{3}}{2-\sqrt{3}}+\sqrt{3}$

$\Rightarrow \frac{1-2\sqrt{3}+2\sqrt{3}-3}{2-\sqrt{3}}$

$\Rightarrow \frac{-2}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}$

$\Rightarrow \frac{-2(2+\sqrt{3})}{4-3}\Rightarrow -2(2+\sqrt{3})$

$\theta =ta{n}^{-1}\left(\frac{1-2\sqrt{3}}{2-\sqrt{3}}\right)$