Maximum Value

We have $f(x)=\sin{2x}+2\sin{x}$ , the function is periodic with period 2 $\pi$ .

we have to differentiate $f(x)$ and then equate it to 0 to find $f_{max}(x)$ .

$f^{'}x=2\cos{2x}+2\cos{x}=0 \\ \implies 2\cos^2{x}+\cos{x}-1=0 \\ \implies \cos{x}=\dfrac{1}2,-1 \implies \sin{x}=\dfrac{\sqrt{3}}{2},0 \\ \text{0 is rejected } \\ f_{Max}=2\times\dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}+2\times\dfrac{\sqrt{3}}{2} \implies \dfrac{3\sqrt{3}}{2} \\ a=b=3,c=2 \\ \text{Ans } \implies \dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{6} \implies\boxed{1}$