Find the global maximum of $2\sin\left(x\right)-\cos\left(2x\right)$

Question

I found local maximum and minimum of function $2\sin\left(x\right)-\cos\left(2x\right)$, but I don't know how to find global maximum.

Answer 1

You can write the following using $\cos(2x) = 1-2\sin^2(x)$ $$f(x) = 2\sin( x) - \cos (2x) = 2 \sin(x) + 2\sin^2(x) -1 \\= 2(\sin(x) +\tfrac{1}{2})^2 - \tfrac{3}{2}$$

From here note that the square term is always positive, maximum occurs at $\sin(x) = 1$ and minimum at $\sin(x) = \tfrac{-1}{2}$.

Thus maximum $= 3$ and minimum = $\tfrac{-3}{2}$.