How do you find exact solutions of \displaystyle{\cos{{2}}}{x}-{\cos{{x}}}={0} in the interval <0,2π) ?
\displaystyle{x}={0},\frac{{{2}\pi}}{{3}},\frac{{{4}\pi}}{{3}} Explanation:Recall that \displaystyle{\cos{{\left({2}{x}\right)}}}={{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}} .Now, we have \displaystyle{{\cos}^{{2}}{x}}-{{\sin}^{{2}}{x}}-{\cos{{x}}}={0} ...

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Bdub Apr 18, 2016 Use the Property:\displaystyle{\cos{{2}}}{A}={2}{{\cos}^{{-{{2}}}}{A}}-{1}\displaystyle{\cos{{x}}}-{\left({2}{{\cos}^{{2}}{x}}-{1}\right)}={0}\displaystyle-{1}{\left<{\cos{{x}}}-{2}{{\cos}^{{2}}{x}}+{1}\right>}={0} ...
Express\displaystyle{\cos{{2}}}{x}in terms of\displaystyle{{\cos}^{{2}}{x}} , then solve the resulting quadratic to find\displaystyle{\cos{{x}}}={1} , hence\displaystyle{x}={2}{n}\pi ...
How do you find all \displaystyle{\cos{{2}}}{x}-{\cos{{6}}}{x}={0} in the interval \displaystyle{\left<{0},{2}\pi\right)} ?
\displaystyle{0},\frac{\pi}{{4}},\frac{\pi}{{2}},\frac{{{3}\pi}}{{4}},\pi,\frac{{{5}\pi}}{{4}},\frac{{{3}\pi}}{{2}},\frac{{{7}\pi}}{{4}} Explanation:Use trig identity: \displaystyle{\cos{{a}}}-{\cos{{b}}}=-{2}{\sin{{\left(\frac{{{a}+{b}}}{{2}}\right)}}}{\sin{{\left(\frac{{{a}-{b}}}{{2}}\right)}}} ...
If \displaystyle{\cos{{\left({2}{x}\right)}}}+{\cos{{x}}}={0} , how would you find the value of \displaystyle{x} on \displaystyle{\left<{0},{2}\pi\right)} ?
\displaystyle{x}=\frac{\pi}{{3}},\frac{{{5}\pi}}{{3}},\pi Explanation:Use the identity\displaystyle{\cos{{2}}}{x}={1}-{2}{{\sin}^{{2}}{x}}to start the solving process. \displaystyle{\cos{{2}}}{x}+{\cos{{x}}}={0} ...

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cos 3x+cos x=0 =>2 cos 2x cos x=0 =>cos 2x=0; or ; cos x=0 For x belong to<-pi,pi>; cos x=0 =>x = pi/2 , -pi/2 , 3 pi/2 , -3 pi/2 ...
2 cos 2x cos x=0 =>cos 2x=0; or ; cos x=0 For x belong to<-pi,pi>; cos x=0 =>x = pi/2 , -pi/2 , 3 pi/2 , -3 pi/2 ..." role="text" class="MathExpression_mathExpression__22QI1 ">
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