Solve the equation $sin x + cos x = 1$

x = (4 n + 1) \frac{π}{4}

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x = (4 n + 1) \frac{π}{2}

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x = (4 n + 1) \frac{π}{8}

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x = (4 n + 1) \frac{π}{16}

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Solution

The correct option is B $x = (4 n + 1) \frac{π}{2}$
Given, $sin x + cos x = 1$
$\Rightarrow \frac{1}{\sqrt{2}} sin x + \frac{1}{\sqrt{2}} cos x = \frac{1}{\sqrt{2}}$
$\Rightarrow sin \frac{π}{4} sin x + cos \frac{π}{4} cos x = \frac{1}{\sqrt{2}}$
$\Rightarrow cos (x - \frac{π}{4}) = cos \frac{π}{4}$
Hence general solutionis given by,
$\Rightarrow x - \frac{π}{4} = 2 n π \pm \frac{π}{4}, n \in Z$
$\Rightarrow x = 2 n π$ or $x = 2 n π + \frac{π}{2} = (4 n + 1) \frac{π}{2}$

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