# I can't figure out how to simplify: \frac 12=\frac{\sin^2 x}{\tan x} What

I can't figure out how to simplify:
$$\displaystyle{\frac{{12}}{=}}{\frac{{{{\sin}^{{2}}{x}}}}{{{\tan{{x}}}}}}$$
What would be my next step to get to $$\displaystyle{x}={45}^{{\circ}}$$?

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Tiefdruckot
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 $$\displaystyle{n}\in{\mathbb{{{Z}}}}$$
###### Not exactly what youâ€™re looking for?
puhnut1m
$$\displaystyle{\frac{{{{\sin}^{{2}}{x}}}}{{{\tan{{x}}}}}}={\frac{{{{\sin}^{{2}}{x}}}}{{{\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}}}}={\sin{{x}}}{\cos{{x}}}$$
$$\displaystyle{\sin{{x}}}{\cos{{x}}}={\frac{{12}}{{\sin{{2}}}}}{x}$$
$$\displaystyle{\frac{{12}}{{\sin{{2}}}}}{x}={\frac{{12}}{\Leftrightarrow}}{\sin{{2}}}{x}={1}\Leftrightarrow{2}{x}={90}^{{\circ}}+{k}{360}^{{\circ}}$$
$$\displaystyle{x}={45}^{{\circ}}+{k}{180}^{{\circ}}$$
Vasquez

cancel $$\sin x$$ on either side, simplify
$$\sin 2x=1 \to 2x=\frac{\pi}{2} \pm 2k \pi \to x=\frac{\pi}{4} \pm k \pi$$