![\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)](/tpl/images/0304/2563/5f1d3.png)
Ineed to evaluate this problem & find its degrees
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