`sin(4x) = -2sin(2x)` Find the exact solutions of the equation in the interval `[0, 2pi).`

Wednesday, May 7, 2014

$\displaystyle{\sin{{\left({4}{x}\right)}}}=-{2}{\sin{{\left({2}{x}\right)}}}$ Find the exact solutions of the equation in the interval $\displaystyle{\left[{0},{2}\pi\right)}.$

By a double angle formula $\displaystyle{\sin{{\left({4}{x}\right)}}}={2}{\sin{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}.$

Therefore our equation may be rewritten as

$\displaystyle{2}{\sin{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}+{2}{\sin{{\left({2}{x}\right)}}}={0},$ or

$\displaystyle{\sin{{\left({2}{x}\right)}}}{\left({\cos{{\left({2}{x}\right)}}}+{1}\right)}={0}.$

So $\displaystyle{\sin{{\left({2}{x}\right)}}}={0}$ or $\displaystyle{\cos{{\left({2}{x}\right)}}}=-{1}.$

The general solutions are $\displaystyle{2}{x}={k}\pi,$ or $\displaystyle{x}=\frac{{{k}\pi}}{{2}},$

and $\displaystyle{2}{x}=-\frac{\pi}{{2}}+{2}{k}\pi,$ so $\displaystyle{x}=-\frac{\pi}{{4}}+{k}\pi,$ where $\displaystyle{k}$ is any integer.

The roots in the interval $\displaystyle{\left[{0},{2}\pi\right)}$ are

$\displaystyle{x}={0},$ $\displaystyle{x}=\frac{\pi}{{2}},$ $\displaystyle{x}=\pi,$ $\displaystyle{x}=\frac{{{3}\pi}}{{2}},$ $\displaystyle{x}=\frac{{{3}\pi}}{{4}},$ $\displaystyle{x}=\frac{{{7}\pi}}{{4}}.$

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Wednesday, May 7, 2014

$\displaystyle{\sin{{\left({4}{x}\right)}}}=-{2}{\sin{{\left({2}{x}\right)}}}$ Find the exact solutions of the equation in the interval $\displaystyle{\left[{0},{2}\pi\right)}.$

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