find all solutions in [0,2pi): 2sin(x)-sin(2x)=0

Answer 1

If we rewrite sin(2x) = 2sin(x)cos(x), we get 2sin(x) - 2sin(x)cos(x) = 0.

Factoring give us sin(x)(1 - cos(x)) = 0; sin(x) = 0, or cos(x) = 1

And we get x = 0 or x = π.

Solution of this equation on interval [0, 2π}: 0 and π

Answer 2

To solve this, start by using this identity:

sin(θ+Ψ)=sin(θ)cos(Ψ)+cos(θ)sin(Ψ)

so sin(2x) = sin(x)cos(x)+cos(x)sin(x)=2sin(x)cos(x)

so 2sin(x)-sin(2x)=2sin(x)-2sin(x)cos(x)=2sin(x)(1-cos(x))

By the way, breaking the expression into factors will be important here:

if

2sin(x)(1-cos(x))=0

that means if either factor is equal to zero that will provide a solution:

one (or possibly more solutions comes from sin(x)=0

if we restrict our solution between (and including) 0 to 2π, we would have 3 solutions from sinx = 0: 0, π and 2π radians.

The other solutions will come from (1-cosx)=0 or cosx = 1; in the same range of x, the solutions would be 0 and 2π, so essentially we have three overall solutions: 0, π and 2π radians.

(I wanted to ask my colleagues to check my work just in case there is something I overlooked)