Question

I need to evaluate this problem & find its degrees

Answer 1

$\bf csc(\theta)=\cfrac{1}{sin(\theta)}\\\\ -----------------------------\\\\ 6sin(\theta)-3csc(\theta)=0\implies 6sin(\theta)-3\cfrac{1}{sin(\theta)}=0 \\\\\\ 6sin(\theta)-\cfrac{3}{sin(\theta)}=0\impliedby \textit{multiplying both sides by }sin(\theta)$

$\bf 6sin^2(\theta)-3=0\implies 6sin^2(\theta)=3\implies sin^2(\theta)=\cfrac{3}{6} \\\\\\ sin(\theta)=\sqrt{\cfrac{1}{2}}\implies sin^{-1}[sin(\theta)]=sin^{-1}\left( \cfrac{1}{\sqrt{2}}\right) \\\\\\ \measuredangle \theta=sin^{-1}\left( \cfrac{1}{\sqrt{2}}\right)\implies \measuredangle \theta=sin^{-1}\left( \cfrac{\sqrt{2}}{2}\right)$

now, if you check your Unit Circle, that's a well known angle for the 1st and 2nd quadrants