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}{√(2)}\right) \\\\\\ \measuredangle \theta=sin^(-1)\left( \cfrac{1}{√(2)}\right)\implies \measuredangle \theta=sin^(-1)\left( \cfrac{√(2)}{2}\right)`

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