`sin(x + pi/6) - sin(x - (7pi)/6) = sqrt(3)/2` Find all solutions of the equation in the interval `[0, 2pi).`

Tuesday, November 17, 2015

$\displaystyle{\sin{{\left({x}+\frac{\pi}{{6}}\right)}}}-{\sin{{\left({x}-\frac{{{7}\pi}}{{6}}\right)}}}=\frac{\sqrt{{{3}}}}{{2}}$ Find all solutions of the equation in the interval $\displaystyle{\left[{0},{2}\pi\right)}.$

Use the formula of difference of sinuses

$\displaystyle{\sin{{\left({a}\right)}}}-{\sin{{\left({b}\right)}}}={2}{\sin{{\left(\frac{{{a}-{b}}}{{2}}\right)}}}{\cos{{\left(\frac{{{a}+{b}}}{{2}}\right)}}}$

and obtain

$\displaystyle{2}{\sin{{\left(\frac{{{2}\pi}}{{3}}\right)}}}{\cos{{\left({x}-\frac{\pi}{{2}}\right)}}}=\frac{\sqrt{{{3}}}}{{2}},$

or $\displaystyle{2}\cdot\frac{\sqrt{{{3}}}}{{2}}\cdot{\cos{{\left({x}-\frac{\pi}{{2}}\right)}}}=\frac{\sqrt{{{3}}}}{{2}},$

or $\displaystyle{\cos{{\left({x}-\frac{\pi}{{2}}\right)}}}=\frac{{1}}{{2}}.$

The general solution is $\displaystyle{x}-\frac{\pi}{{2}}=\pm\frac{\pi}{{3}}+{2}{k}\pi,$ or $\displaystyle{x}=\frac{\pi}{{2}}\pm\frac{\pi}{{3}}+{2}{k}\pi.$

The solutions on the interval $\displaystyle{\left[{0},{2}\pi\right)}$ are $\displaystyle{x}=\frac{{{5}\pi}}{{6}}$ and $\displaystyle{x}=\frac{\pi}{{6}}.$

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Tuesday, November 17, 2015

$\displaystyle{\sin{{\left({x}+\frac{\pi}{{6}}\right)}}}-{\sin{{\left({x}-\frac{{{7}\pi}}{{6}}\right)}}}=\frac{\sqrt{{{3}}}}{{2}}$ Find all solutions of the equation in the interval $\displaystyle{\left[{0},{2}\pi\right)}.$

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