Question

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

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

Answer 1

The correct option is B $2n\pi +\frac{3\pi }{4}$

Given that, $\sin x-\cos x=\sqrt{2}$
$\Rightarrow \frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x=1$
$\sin {45}^{\circ }\sin x-\cos {45}^{\circ }\cos x=1$
$\Rightarrow \cos (x+\frac{\pi }{4})=-1$
$\Rightarrow \cos (x+\frac{\pi }{4})=\cos \left(\pi \right)$
$\Rightarrow x+\frac{\pi }{4}=2n\pi +\pi$
$\Rightarrow x=2n\pi +\frac{3\pi }{4}$