Soviet Sine

If $3sin(2α) - 2sin(2β) = 0$ , that means $3sin(2α) = 2sin(2β)$ . Using the double angle identity, $3sin(α)cos(α) = 2sin(β)cos(β)$ . Now, let's square both sides. This gives us $9sin(α)^2cos(α)^2 = 4sin(β)^2cos(β)^2$ and we know that $sin(α)^2 + cos(α)^2 = 1$ and $sin(β)^2 + cos(β)^2 = 1$ by the Pythagorean identity. Thus, $9sin(α)^2(1-sin(α)^2) = 4sin(β)^2(1-sin(β)^2)$ . By the first equation, we know that $3sin(α)^2 + 2sin(β)^2 = 1$ , so we can use a linear system of equations with $sin(α)^2$ and $sin(β)^2$ . Once this is done, $sin(α)^2 = \frac{1}{9}$ and $sin(β)^2 = \frac{1}{3}$ . Since we want to find $α+2β$ , we can use the values of $sin(α)$ and $sin(β)$ , the pythagorean identities to derive $cos(α)$ and $cos(β)$ , and the double angle identities to derive $cos(2α)$ and $cos(2β)$ to derive this quantity. Using these identities, $sin(α+2β) = sin(α)cos(2β) + cos(α)sin(2β) = 1$ , so then $α+2β = π/2$ . Thus, $a+b = 1 + 2 = 3$ which is the final answer.