Solve 2 sin(y) = sin(x) + sin(x + y) for x and y in [0, 2pi).

Step-by-step solution

1. 1

   Convert the right-hand sum into a product. The identity $\sin A+\sin B=2\sin\tfrac{A+B}{2}\cos\tfrac{A-B}{2}$ with $A=x+y$ and $B=x$ gives

   $\sin x+\sin(x+y)=2\sin\!\left(x+\frac y2\right)\cos\frac y2$

2. 2

   Rewrite the left-hand side with the double-angle formula.

   $2\sin y=4\sin\frac y2\cos\frac y2$

   Both sides now carry the common factor $\cos\tfrac y2$, which is the structural key to the problem.

3. 3

   Factor rather than divide. Moving everything to one side:

   $2\cos\frac y2\left[2\sin\frac y2-\sin\!\left(x+\frac y2\right)\right]=0$

   Dividing both sides by $\cos\tfrac y2$ instead of factoring would silently discard every solution with $\cos\tfrac y2=0$ — the most common way this problem is answered incompletely.

4. 4

   Branch 1: $\cos\tfrac y2=0$. On $[0,2\pi)$ this means $\tfrac y2=\tfrac\pi2$, so

   $y=\pi\qquad\text{for every }x\in[0,2\pi)$

   Check it directly: $2\sin\pi=0$ and $\sin x+\sin(x+\pi)=\sin x-\sin x=0\;\checkmark$. This is a whole line of solutions, not a single point.

5. 5

   Branch 2: $\sin\!\left(x+\tfrac y2\right)=2\sin\tfrac y2$. A sine is bounded by $1$, so this branch requires

   $\left|2\sin\frac y2\right|\le 1\iff\left|\sin\frac y2\right|\le\frac12\iff y\in\left[0,\frac\pi3\right]\cup\left[\frac{5\pi}{3},2\pi\right)$

   and for each admissible $y$ the matching $x$ values are

   $x\equiv\arcsin\!\left(2\sin\frac y2\right)-\frac y2\quad\text{or}\quad x\equiv\pi-\arcsin\!\left(2\sin\frac y2\right)-\frac y2\pmod{2\pi}$

6. 6

   Summarise the full solution set. The solutions are the horizontal line $y=\pi$ together with the two curves of Branch 2. Numerical spot checks confirm both families: at $y=\pi$ with $x=0.3,\,1.234,\,2.7,\,5.5$ the residual is below $4\times 10^{-16}$, and Branch 2 at $y=0$ forces $\sin x=0$, giving $x=0$ and $x=\pi$, which satisfy the original equation exactly.