Solve this: abs(2cos3x)=1|2cos3x|=1 ---> (x=... + ...)?

abs(2cos3x)=1|2cos3x|=1

Will it be: x=pi/9+2/3kpix=π9+23kπ AND x=-pi/9+2/3kpix=π9+23kπ
OR ......
x=pi/9+1/3kpix=π9+13kπ AND x=-pi/9+1/3kpix=π9+13kπ

1 Answer
Apr 2, 2017

x=2/3kpi+-pi/9x=23kπ±π9 and x=2/3kpi+-(2pi)/9x=23kπ±2π9

Explanation:

As |2cos3x|=1|2cos3x|=1, we have

either 2cos3x=12cos3x=1 i.e. cos3x=1/2=cos(pi/3)cos3x=12=cos(π3)

and 3x=2kpi+-pi/33x=2kπ±π3 or x=2/3kpi+-pi/9x=23kπ±π9

or 2cos3x=-12cos3x=1 i.e. cos3x=-1/2=cos((2pi)/3)cos3x=12=cos(2π3)

and 3x=2kpi+-(2pi)/33x=2kπ±2π3 or x=2/3kpi+-(2pi)/9x=23kπ±2π9