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

Jaya Legge 2021-08-16 Answered
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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Expert Answer

dieseisB
Answered 2021-08-17 Author has 12897 answers
\(\displaystyle{\sin{{2}}}{x}={2}{\sin{{x}}}{\cos{{x}}}\)
\(\displaystyle{\sin{{x}}}{\cos{{x}}}= {\left({\frac{{{1}}}{{{2}}}}\right)}{\sin{{2}}}{x}={\frac{{{1}}}{{{2}}}}\Rightarrow{\sin{{2}}}{x}={1}\)
\(\displaystyle{\sin{{2}}}{x}={1}\Rightarrow{\arcsin{{2}}}{x}={\frac{{\pi}}{{{4}}}}\Rightarrow{x}={\frac{{\pi}}{{{4}}}}\)
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!
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