If the equation 2 sin^2 x + frac{sin 2x}{2} = | vedclass

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(B) Given the equation $2 \sin^2 x + \frac{\sin 2x}{2} = k$.Using the identity $2 \sin^2 x = 1 - \cos 2x$,we get $1 - \cos 2x + \frac{\sin 2x}{2} = k$

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$3$

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(B) Given the equation $2 \sin^2 x + \frac{\sin 2x}{2} = k$.Using the identity $2 \sin^2 x = 1 - \cos 2x$,we get $1 - \cos 2x + \frac{\sin 2x}{2} = k$.Rearranging,we have $\frac{1}{2} \sin 2x - \cos 2x = k - 1$.The expression $a \sin 	heta + b \cos 	heta$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$a = \frac{1}{2}$ and $b = -1$,so $\sqrt{a^2 + b^2} = \sqrt{\frac{1}{4} + 1} = \frac{\sqrt{5}}{2}$.Thus,$-\frac{\sqrt{5}}{2} \leq k - 1 \leq \frac{\sqrt{5}}{2}$.Adding $1$ to all sides,$1 - \frac{\sqrt{5}}{2} \leq k \leq 1 + \frac{\sqrt{5}}{2}$.Since $\sqrt{5} \approx 2.236$,we have $1 - 1.118 \leq k \leq 1 + 1.118$,which is $-0.118 \leq k \leq 2.118$.The integral values of $k$ are $0, 1, 2$.The sum of these values is $0 + 1 + 2 = 3$.

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

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