Here's how I approached the problem.

$f'(x) = -\sin x - 2a$

$f'(x) \geq 0$
$\Rightarrow -\sin x -2a \geq 0$

$\Rightarrow 2a \leq -\sin x$

$\Rightarrow 2a+1 \leq 1- \sin x \tag{\*}$

Since range of $\sin x$ is $[-1,1]$,
$\Rightarrow 2a+1\leq 1-(-1)$

$\Rightarrow 2a+1\leq 2$

Hence maximum value of $2a+1$ is coming to be $2$. However the answer to the question is $0$. What am I doing wrong?