\(\displaystyle{\cos{{\left({2}{x}\right)}}}={2}{{\cos}^{{2}}{\left({x}\right)}}-{1}\)
\(\displaystyle{2}{{\cos}^{{2}}{\left({x}\right)}}-{1}={\cos{{\left({x}\right)}}}\)
\(\displaystyle{2}{{\cos}^{{2}}{\left({x}\right)}}-{1}-{\cos{{\left({x}\right)}}}={0}\)
\(\displaystyle{2}{{\cos}^{{2}}{\left({x}\right)}}-{\cos{{\left({x}\right)}}}-{1}={0}\)
\(\displaystyle{\left({2}{\cos{{\left({x}\right)}}}+{1}\right)}{\left({\cos{{\left({x}\right)}}}-{1}\right)}={0}\)
\(\displaystyle{2}{\cos{{\left({x}\right)}}}+{1}={0}\)
\(\displaystyle{2}{\cos{{\left({x}\right)}}}=-{1}\)
\(\displaystyle{\cos{{\left({x}\right)}}}=-\frac{{1}}{{2}}\)
\(\displaystyle{x}={{\cos}^{{-{{1}}}}{\left(-\frac{{1}}{{2}}\right)}}\)
\(x = \pi/3 [ 2\pi - \pi/3 = 5\pi/3 ] \)
\(x = \pi/3\ and\ 5 \pi/3\)
cos is positive in the first and fourth quadrant.
\(\displaystyle{\cos{{\left({x}\right)}}}-{1}={0}\)
\(\displaystyle{\cos{{\left({x}\right)}}}={1}\)
\(\displaystyle{x}={{\cos}^{{-{{1}}}}{\left({1}\right)}}\)
\(\displaystyle{x}={0},{2}\pi\)
Find the general solution to cos2x=-cosx
Find the general solution to cos2x=-cosx
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Answers (1)
2021-02-12
Relevant Questions
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To solve this, the first thing I did was find the general solutionto the homogeneous equivalent, and got
\(\displaystyle{c}_{{1}}{e}^{{-{5}\frac{{x}}{{2}}}}+{c}_{{2}}{e}^{{{3}\frac{{x}}{{2}}}}\)
Then i used the form \(\displaystyle{K}{\cos{{\left({w}{x}\right)}}}+{M}{\sin{{\left({w}{x}\right)}}}\) and got \(\displaystyle-{2.72}{\cos{{\left({5}{x}\right)}}}+{2.56}{\sin{{\left({5}{x}\right)}}}\) as a solution of the nonhomogeneous ODE
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