You're on the right track!

what you need to do is to move the 1000 to the RHS to get an equation to equal 0.

then let f(x)=(x^2)+10sinx-1000

Then all you need to do is to find a value of x for when f(x)<0, and a value of x for when f(x)>0 (you can find "nice" values of x to make the sin function easy to calculate; try x=0 for starters) - proving that there is a solution between these two points, hence this is where the IVT comes in.