How do you solve the following equation $2cos3x=1$ in the interval [0, pi]?

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Answer 1 · 2016-07-03T16:56:38

Values that $x$ can take in interval are $[0,2pi]$ are ${pi/9,(5pi)/9,(7pi)/9,(11pi)/9,(13pi)/9,(17pi)/9}$

As $2cos3x=1$ , we have $cos3x=1/2=cos(pi/3)$

Hence $3x=2npi+-pi/3$ , where $n$ is an integer.

Hence, $3x$ can take values ${......,-(5pi)/3,-pi/3,pi/3,(5pi)/3,(7pi)/3,(11pi)/3,(13pi)/3,(17pi)/3,(19pi)/3,....}$

and $x$ can take values ${.......,-(5pi)/9,-pi/9,pi/9,(5pi)/9,(7pi)/9,(11pi)/9,(13pi)/9,(17pi)/9,(19pi)/9,....}$

and values that $x$ can take in interval are $[0,2pi]$ are ${pi/9,(5pi)/9,(7pi)/9,(11pi)/9,(13pi)/9,(17pi)/9}$

How do you solve the following equation 2cos3x=12cos3x=1 in the interval [0, pi]?