Question #12b1c

1 Answer
Dec 21, 2017

The solutions are k/18pi, where k in {0,5,7,17,18,19,29,31}

Explanation:

2tan(x)cos(3x)=-sqrt(3)tan(x)

Move everything to one side
2tan(x)cos(3x)+sqrt(3)tan(x)=0

Factor tan(x) out
tan(x)(2cos(3x)+sqrt(3))=0

By zero product property, we have
tan(x)=0 or 2cos(3x)+sqrt(3)=0

tan(x)=0 has solutions x=kpi, where k is an integer.
Since x in [0,2pi), only 0 and pi are solutions.

Moving on with the other term
2cos(3x)+sqrt(3)=0
cos(3x)+sqrt(3)/2=0
cos(3x)=-sqrt(3)/2
Here a graph of cos(x) would be helpful - it gives a little suggestion (click and drag for (x,y) values and special points)
graph{(y-cos(x))(y+sqrt(3)/2)=0 [-1,7,-2,2]}
We need to recall, that sqrt(3)/2=cos(pi/6)
From the graph we see, that the solution is 3x=2kpi+pi pm pi/6 or x=((2k+1)/3pm1/18)pi
Since x in [0,2pi), only solutions are
(1/3-1/18)pi,(1/3+1/18)pi,(3/3-1/18)pi,(3/3+1/18)pi,(5/3-1/18)pi,(5/3+1/18)pi
or simplified
5/18pi,7/18pi,17/18pi,19/18pi,29/18pi,31/18pi