First we need to check that the identity is true. Let x=0, then sin(5x)=0, so the right-hand expression should also evaluate to zero:
1(0+2)-0-1=1
The identity is false. Also the maximum of sin(5x) is 1 and its minimum is -1, whereas the maximum of the right-hand expression exceeds 1 and its minimum is smaller than that of sin(5x).
However, when x=π/2:
sin(5π/2)=1=0-(-1)-0, so x=π/2 is a solution if the identity is converted into an equation.
When x=π/6:
sin(5π/6)=½=¾(1+0)-¼-0, so x=π/6 is also a solution. There are many other solutions.