Question

solve the following equation for all values of x: sin^2x+sinxcosx

Trigonometric equation and identitie
ANSWERED
asked 2021-03-08
solve the following equation for all values of x: \(\displaystyle{{\sin}^{{2}}{x}}+{\sin{{x}}}{\cos{{x}}}\)

Answers (1)

2021-03-09
\(\displaystyle{\sin{{\left({x}\right)}}}{\left({\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}\right)}\). If this expression equals zero, \(\displaystyle{\sin{{\left({x}\right)}}}={0}\) or \(\displaystyle{\sin{{\left({x}\right)}}}=-{\cos{{\left({x}\right)}}}\), so \(\displaystyle{\tan{{\left({x}\right)}}}=-{1}.\)
Solutions: \(\displaystyle{\sin{{\left({x}\right)}}}={0}:{x}={n}{\left(\pi\right)},{\tan{{\left({x}\right)}}}=-{1}:{x}={\left({4}{n}-{1}\right)}\frac{{\pi}}{{4}}\) where n is an integer. x is in radians. To convert to degrees put \(\displaystyle{\left(\pi\right)}={180}\): 0, 180, 360, ..., 135, 315, ... for example.
\(\displaystyle{\cos{{\left({2}{x}\right)}}}={1}-{2}{{\sin}^{{2}}{\left({x}\right)}}\), so \(\displaystyle{{\sin}^{{2}}{\left({x}\right)}}=\frac{{{1}-{\cos{{\left({2}{x}\right)}}}}}{{2}}\)
\(\displaystyle{\sin{{\left({2}{x}\right)}}}={2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}\), so \(\displaystyle{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}={\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{x}\right)}}}\)
\(\displaystyle{{\sin}^{{2}}{\left({x}\right)}}+{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}={\left(\frac{{1}}{{2}}\right)}{\left({1}-{\cos{{\left({2}{x}\right)}}}+{\sin{{\left({2}{x}\right)}}}\right)}\). Above solutions apply.
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