Let \(\displaystyle{t}\:={\tan{{x}}}\). Since \(\displaystyle{\frac{{12}}{=}}{t}{{\sin}^{{2}}{x}}={t}^{{3}}{{\cos}^{{2}}{x}}={\frac{{{t}^{{3}}}}{{{1}+{t}^{{2}}}}}\)

\(\displaystyle{0}={2}{t}^{{3}}-{t}^{{2}}-{1}={\left({t}-{1}\right)}{\left({2}{t}^{{2}}+{t}+{1}\right)}\)

Since \(\displaystyle{2}{t}^{{2}}+{t}+{1}{>}{0},{t}={1}\). This is achieved when \(\displaystyle{x}={\left({45}+{180}{n}\right)}^{{\circ}}\) for \(\displaystyle{n}\in{\mathbb{{{Z}}}}\)

\(\displaystyle{0}={2}{t}^{{3}}-{t}^{{2}}-{1}={\left({t}-{1}\right)}{\left({2}{t}^{{2}}+{t}+{1}\right)}\)

Since \(\displaystyle{2}{t}^{{2}}+{t}+{1}{>}{0},{t}={1}\). This is achieved when \(\displaystyle{x}={\left({45}+{180}{n}\right)}^{{\circ}}\) for \(\displaystyle{n}\in{\mathbb{{{Z}}}}\)