The general solution of sin x - cos x = sqrt | vedclass

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Question

(b) $\sin x\frac{1}{{\sqrt 2 }} - \cos x\frac{1}{{\sqrt 2 }} = 1$

$\Rightarrow \cos \left( {x + \frac{\pi }{4}} \right) = - 1$

==> $x + \frac

Vedclass · Accepted Answer

$2n\pi + \frac{{3\pi }}{4}$

Vedclass · Answer

(b) $\sin x\frac{1}{{\sqrt 2 }} - \cos x\frac{1}{{\sqrt 2 }} = 1$

$\Rightarrow \cos \left( {x + \frac{\pi }{4}} \right) = - 1$

==> $x + \frac{\pi }{4} = 2n\pi \pm \pi $

$\Rightarrow 2n\pi + \frac{{3\pi }}{4}{\rm{\,\,or\,\,}}\,\,2n\pi - \frac{{5\pi }}{4}$ .

The general solution of $\sin x - \cos x = \sqrt 2 $, for any integer $n$ is

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