solve the equation:
root3*sin2$\displaystyle \theta$+2sin^2$\displaystyle \theta$=1 for 0<$\displaystyle \theta$<$\displaystyle \pi$
thanks!
I'll nudge you in the right direction:
$\displaystyle \sqrt{3}\sin(2\vartheta)+2\sin^2\vartheta=1$
Note that $\displaystyle \cos(2\vartheta)=1-2\sin^2\vartheta$.
So we end up with $\displaystyle \sqrt{3}\sin(2\vartheta)=1-2\sin^2\vartheta\implies \sqrt{3}\sin(2\vartheta)=\cos(2\vartheta)$
Thus we end up with $\displaystyle \sqrt{3}=\cot(2\vartheta)$
Can you take it from here?
--Chris