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

Donald Johnson 2022-01-02 Answered
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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Expert Answer

Tiefdruckot
Answered 2022-01-03 Author has 2049 answers
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 \(\displaystyle{n}\in{\mathbb{{{Z}}}}\)
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puhnut1m
Answered 2022-01-04 Author has 2383 answers
\(\displaystyle{\frac{{{{\sin}^{{2}}{x}}}}{{{\tan{{x}}}}}}={\frac{{{{\sin}^{{2}}{x}}}}{{{\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}}}}={\sin{{x}}}{\cos{{x}}}\)
\(\displaystyle{\sin{{x}}}{\cos{{x}}}={\frac{{12}}{{\sin{{2}}}}}{x}\)
\(\displaystyle{\frac{{12}}{{\sin{{2}}}}}{x}={\frac{{12}}{\Leftrightarrow}}{\sin{{2}}}{x}={1}\Leftrightarrow{2}{x}={90}^{{\circ}}+{k}{360}^{{\circ}}\)
\(\displaystyle{x}={45}^{{\circ}}+{k}{180}^{{\circ}}\)
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Vasquez
Answered 2022-01-09 Author has 8850 answers

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

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