The sum of all x in [0, pi] which satisfy the | vedclass

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(C) Given equation: $\sin x + \frac{1}{2} \cos x = \sin^2(x + \frac{\pi}{4})$Multiply by $2$: $2 \sin x + \cos x = 2 \sin^2(x + \frac{\pi}{4})$Using t

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$\pi$

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(C) Given equation: $\sin x + \frac{1}{2} \cos x = \sin^2(x + \frac{\pi}{4})$Multiply by $2$: $2 \sin x + \cos x = 2 \sin^2(x + \frac{\pi}{4})$Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:$2 \sin x + \cos x = 2 [\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}]^2$$2 \sin x + \cos x = 2 [\frac{1}{\sqrt{2}}(\sin x + \cos x)]^2$$2 \sin x + \cos x = 2 	imes \frac{1}{2}(\sin x + \cos x)^2$$2 \sin x + \cos x = \sin^2 x + \cos^2 x + 2 \sin x \cos x$Since $\sin^2 x + \cos^2 x = 1$:$2 \sin x + \cos x = 1 + 2 \sin x \cos x$$2 \sin x - 2 \sin x \cos x + \cos x - 1 = 0$$2 \sin x(1 - \cos x) - 1(1 - \cos x) = 0$$(2 \sin x - 1)(1 - \cos x) = 0$This gives $\sin x = \frac{1}{2}$ or $\cos x = 1$.For $x \in [0, \pi]$,$\sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}$.For $x \in [0, \pi]$,$\cos x = 1 \Rightarrow x = 0$.The sum of all such $x$ is $\frac{\pi}{6} + \frac{5\pi}{6} + 0 = \pi$.

The sum of all $x \in [0, \pi]$ which satisfy the equation $\sin x + \frac{1}{2} \cos x = \sin^2(x + \frac{\pi}{4})$ is

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