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(A) Given curve: $y = \sqrt{2} \sin \left(2x + \frac{\pi}{4}\right)$.First,find the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = \sqrt{2} \cos \left(2x

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$2x + y - \frac{\pi}{2} - 1 = 0$

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(A) Given curve: $y = \sqrt{2} \sin \left(2x + \frac{\pi}{4}\right)$.First,find the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = \sqrt{2} \cos \left(2x + \frac{\pi}{4}\right) \cdot 2 = 2\sqrt{2} \cos \left(2x + \frac{\pi}{4}\right)$.At $x = \frac{\pi}{4}$,the slope $m$ is:$m = \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = 2\sqrt{2} \cos \left(2 \cdot \frac{\pi}{4} + \frac{\pi}{4}\right) = 2\sqrt{2} \cos \left(\frac{3\pi}{4}\right) = 2\sqrt{2} \left(-\frac{1}{\sqrt{2}}\right) = -2$.Now,find the $y$-coordinate at $x = \frac{\pi}{4}$:$y = \sqrt{2} \sin \left(2 \cdot \frac{\pi}{4} + \frac{\pi}{4}\right) = \sqrt{2} \sin \left(\frac{3\pi}{4}\right) = \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = 1$.The point of tangency is $\left(\frac{\pi}{4}, 1\right)$.The equation of the tangent is $(y - y_1) = m(x - x_1)$:$(y - 1) = -2 \left(x - \frac{\pi}{4}\right)$$y - 1 = -2x + \frac{\pi}{2}$$2x + y - \frac{\pi}{2} - 1 = 0$.

The equation of the tangent to the curve $y = \sqrt{2} \sin \left(2x + \frac{\pi}{4}\right)$ at $x = \frac{\pi}{4}$ is:

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