`tan(2x) - 2cos(x) = 0` Find the exact solutions of the equation in the interval [0, 2pi).

`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 identity`cos(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`