calculating 2 constants in a function M={f∈C[0,2π],∫02πf(x)sin⁡x,dx=π,∫02πf(x)sin⁡2x,dx=2π} a,b∈R,g∈M,g(x)=asin⁡x+bsin⁡2x,x∈[0,2π] Ive

The overall pattern is
${\displaystyle {\int }_0^{2\pi }\sin \left(mx\right)\sin \left(nx\right)dx=\pi {\delta }_{mn}}$
where ${\displaystyle {\delta }_{mn}}$ is 1 when m=n and 0 otherwise.
The average value of ${\displaystyle {\sin }^2\text{ is }\frac{1}{2}}$ and sines with different periods are orthogonal over a common period. You can prove the the first from ${\displaystyle {\sin }^{2}x=\frac{12}{-}\frac{12}{\cos \left(2x\right)}}$ The ${\displaystyle \cos \left(2x\right)}$ integrates away. You can prove the first by using the function product identities, which gets you a sum of sine waves that integrate to zero as well. The same thing works for cosines.
This works because the sines and cosines form an orthogonal basis for the ${\displaystyle 2\pi }$ periodic functions and you are effectively expanding a function in that basis.