Question

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

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

Answer 1

The correct option is B

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

Find the general solution

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

Dividing by $\sqrt{2}$ on both sides,

$$\begin{gathered} \Rightarrow \frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x=1 \\ \Rightarrow \sin \frac{\mathrm{\pi }}{4}\sin x-cos\frac{\mathrm{\pi }}{4}\cos x=1\left[\because \sin \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}},\cos \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}\right] \\ \Rightarrow -\cos \left(x+\frac{\mathrm{\pi }}{4}\right)=1\left[\because \cos \left(A\right)\cos \left(B\right)-\sin \left(A\right)\sin \left(B\right)=\cos \left(A+B\right)\right] \\ \Rightarrow \cos \left(x+\frac{\mathrm{\pi }}{4}\right)=-1 \\ \Rightarrow \cos \left(x+\frac{\mathrm{\pi }}{4}\right)=\mathrm{cos\pi } \end{gathered}$$

Therefore, $x+\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }+\mathrm{\pi }$

$$\begin{gathered} \Rightarrow \mathrm{x}=2\mathrm{n\pi }+\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }+3\frac{\mathrm{\pi }}{4} \end{gathered}$$

Hence, option (B) is the correct answer.