Question

If the equation $2{\sin }^{2}x+\frac{\sin 2x}{2}=k$ has atleast one real solution, then the sum of all integral value of $k$ is

Answer 1

$2{\sin }^{2}x+\frac{\sin 2x}{2}=k\Rightarrow -\cos 2x+\frac{\sin 2x}{2}=k-1\Rightarrow \frac{-\sqrt{5}}{2}\le k-1\le \frac{\sqrt{5}}{2}\Rightarrow \frac{-\sqrt{5}}{2}+1\le k\le \frac{\sqrt{5}}{2}+1\Rightarrow \text{sum}=0+1+2=3$