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

Monday, August 25, 2014

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

You need to evaluate the solution to the equation $\displaystyle{\cos{{2}}}{x}+{\sin{{x}}}={0}$ , hence, you need to use the formula of double angle for $\displaystyle{\cos{{2}}}{x}$ , such that:

$\displaystyle{1}-{2}{{\sin}}^{{2}}{x}+{\sin{{x}}}={0}$

You need to re-arrange the terms, such that:

$\displaystyle-{2}{{\sin}}^{{2}}{x}+{\sin{{x}}}+{1}={0}$

$\displaystyle{2}{{\sin}}^{{2}}{x}-{\sin{{x}}}-{1}={0}$

Yo need to replace t for sin x, such that:

$\displaystyle{2}{{t}}^{{2}}-{t}-{1}={0}$

Using quadratic formula, yields:

$\displaystyle{t}_{{1}}={\left({1}+\frac{\sqrt{{{1}+{8}}}}{{4}}\right)}\Rightarrow{t}_{{1}}=\frac{{{1}+{3}}}{{4}}\Rightarrow{t}_{{1}}={1}$

$\displaystyle{t}_{{2}}={\left({1}-\frac{\sqrt{{{1}+{8}}}}{{4}}\right)}\Rightarrow{t}_{{2}}=\frac{{{1}-{3}}}{{4}}\Rightarrow{t}_{{2}}=-\frac{{1}}{{2}}$

You need to replace sin x for t such that:

$\displaystyle{\sin{{x}}}={1}\Rightarrow{x}=\frac{\pi}{{2}}$

$\displaystyle{\sin{{x}}}=-\frac{{1}}{{2}}\Rightarrow{x}=\pi+\frac{\pi}{{6}}\Rightarrow{x}=\frac{{{7}\pi}}{{6}}$

$\displaystyle{\sin{{x}}}=-\frac{{1}}{{2}}\Rightarrow{x}={2}\pi-\frac{\pi}{{6}}\Rightarrow{x}=\frac{{{11}\pi}}{{6}}$

Hence, the solution to the equation, in $\displaystyle{\left[{0},{2}\pi\right)},$ are $\displaystyle{x}=\frac{\pi}{{2}},{x}=\frac{{{7}\pi}}{{6}},{x}=\frac{{{11}\pi}}{{6}}.$

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Monday, August 25, 2014

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

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