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

1 Answer
Shwetank Mauria
Jul 3, 2016

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

Explanation:

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}#

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