Question

The general solution of $\sin x-\cos x=\sqrt{2}$, for any integers 'n' is_______

• A. $2n\pi +\frac{3\pi }{4}$
• B. $n\pi$
• C. $(2n+1)\pi$
• D. $2n\pi$

Answer 1

Answer: (A)

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

Given, $\sin x-\cos x=\sqrt{2}$,

Divide both sides by ,$\sqrt{2}$

$\Rightarrow \frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x=1$

$\Rightarrow \sin \frac{\pi }{4}\sin x-\cos x\cos \frac{\pi }{4}=1$

$\Rightarrow \cos (\frac{\pi }{4}+x)=-1=\cos \pi$

Thus general solution is given by,

$\frac{\pi }{4}+x=2n\pi \pm \pi$

$\Rightarrow x=2n\pi +\frac{3\pi }{4},2n\pi -\frac{5\pi }{4}$, where $n\in Z$