Precalculus, Chapter 5, 5.5, Section 5.5, Problem 14

tan(2x)-2cos(x)=0, 0<=x<=2pi
tan(2x)-2cos(x)=0
sin(2x)/cos(2x)-2cos(x)=0
sin(2x)-2cos(2x)cos(x)=0
2sin(x)cos(x)-2cos(2x)cos(x)=0
2cos(x)(sin(x)-cos(2x))=0
using the identitycos(2x)=1-2sin^2(x),
2cos(x)(sin(x)-(1-2sin^2(x)))=0
2cos(x)(sin(x)-1+2sin^2(x))=0
solving each part separately,
cos(x)=0
General solutions are,
x=pi/2+2pin , x=(3pi)/2+2pin
Solutions for the range 0<=x<=2pi are,
x=pi/2 , x=(3pi)/2
2sin^2(x)+sin(x)-1=0
Let sin(x)=y
2y^2+y-1=0
solve using the quadratic formula,
y=(-1+-sqrt(1^2-4*2*(-1)))/(2*2)
y=-1,1/2
substitute back y=sin(x)
sin(x)=-1 , sin(x)=1/2
For sin(x)=-1
General solutions are,
x=(3pi)/2+2pin
Solutions for the range 0<=x<=2pi are,
x=(3pi)/2
For sin(x)=1/2
General solutions are,
x=pi/6+2pin , x=(5pi)/6+2pin
solutions for the range 0<=x<=2pi are,
x=pi/6 , (5pi)/6
Combine all the solutions,
x=pi/2 , (3pi)/2 , pi/6 , (5pi)/6