I am trying to find a solution for the problem:
$$2\cdot\sin x \cdot \sin (50°+2x)=\sin (50°), \quad x\in \left[0,\frac{\pi}{2}\right]$$
Approach:
I can check, by inspection, that $x=50°$ is a solution. I have tried open $\sin(50º+2x)$ but it didn't work. I also tried sum product relation but I got nothing interesting. Any hint?
P.S: I'm trying to solve using standard approach because the problem is in a high school level. That means I'm not using calculus.
Using Prosthaphaeresis Formula $$ \sin(10°)+\sin(50°)=2\sin(30°)\cos(20°)=\cos(20°) $$ Hence, we can see that : $$ \sin(50°)=\cos(20°)-\sin(10°) \tag{1}\label{1} $$ Manipulating \eqref{1} :
We know that $\cos(180°+\theta)=-\cos(\theta)$. Hence, $\cos(20°)=-\cos(200°)$.
We know that $\cos(90°+\theta)=-\sin(\theta)$. Hence, $\sin(10°)=-\cos(100°)$
Substituting the values, we get : $$ \sin(50°)=\cos(100°)-\cos(200°) $$ According to Prosthaphaeresis Formula (Reverse Application): $$ 2\sin(A)\sin(B)=\cos(A-B)-\cos(A+B) \\ \implies 2\sin(x)\sin(50°+2x)=\cos(50°+x)-\cos(50°+3x) $$
Since : $$ 2\cdot\sin x \cdot \sin (50°+2x)=\sin (50°) \\ \implies \cos(50°+x)-\cos(50°+3x)=\sin (50°)=\cos(100°)-\cos(200°) \\ \implies \cos(50°+x)-\cos(50°+3x)=\cos(100°)-\cos(200°) $$ Hence, we can see that $x=50°$ is an obvious solution. I think the second solution can also be figured out similarly although I don't think it will be as easy.
