The general solution of sin x + cos x = min_{ | vedclass

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(D) First,we evaluate the minimum value of the expression $f(a) = a^2 - 4a + 6$. Completing the square,we get $f(a) = (a-2)^2 + 2$. The minimum value

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$n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}$

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(D) First,we evaluate the minimum value of the expression $f(a) = a^2 - 4a + 6$. Completing the square,we get $f(a) = (a-2)^2 + 2$. The minimum value of $(a-2)^2 + 2$ is $2$ when $a=2$. Thus,$\min_{a \in \mathbb{R}} \{1, a^2 - 4a + 6\} = \min \{1, 2\} = 1$. Now,we solve the equation $\sin x + \cos x = 1$. Dividing both sides by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$. This simplifies to $\sin(x + \frac{\pi}{4}) = \sin(\frac{\pi}{4})$. The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$. Therefore,$x + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$. Solving for $x$,we get $x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}$.

The general solution of $\sin x + \cos x = \min_{a \in \mathbb{R}} \{1, a^2 - 4a + 6\}$ is

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