Question

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

Answer 1

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}`