# How do you solve \sin x\cos x =\frac{1}{2} for x in the interval [0,2

How do you solve $$\displaystyle{\sin{{x}}}{\cos{{x}}}={\frac{{{1}}}{{{2}}}}$$ for x in the interval $$\displaystyle{\left[{0},{2}\pi\right)}$$?

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dieseisB
$$\displaystyle{\sin{{2}}}{x}={2}{\sin{{x}}}{\cos{{x}}}$$
$$\displaystyle{\sin{{x}}}{\cos{{x}}}= {\left({\frac{{{1}}}{{{2}}}}\right)}{\sin{{2}}}{x}={\frac{{{1}}}{{{2}}}}\Rightarrow{\sin{{2}}}{x}={1}$$
$$\displaystyle{\sin{{2}}}{x}={1}\Rightarrow{\arcsin{{2}}}{x}={\frac{{\pi}}{{{4}}}}\Rightarrow{x}={\frac{{\pi}}{{{4}}}}$$
Check:$$\displaystyle{\sin{{\frac{{\pi}}{{{4}}}}}}{\cos{{\frac{{\pi}}{{{4}}}}}}={\left({\frac{{\sqrt{{{2}}}}}{{{2}}}}\right)}{\left({\frac{{\sqrt{{{2}}}}}{{{2}}}}\right)}={\frac{{{2}}}{{{4}}}}={\frac{{{1}}}{{{2}}}}$$ Perfect!