Question

The general solution of the equation $\sin \theta +\cos \theta =1$ is

• A. $\theta =2n\pi +\frac{\pi }{2},n=0,\pm 1,\pm 2$
• B. $\theta =n\pi +\{{(-1)}^n+1\}\pi /4,n=0,\pm 1,\pm 2$
• C. $\theta =n\pi +\{{(-1)}^n-1\}\frac{\pi }{4},n=0,\pm 1,\pm 2$
• D. $\theta =2n\pi ,n=0,\pm 1,\pm 2+...$

Answer 1

The correct option is C $\theta =n\pi +\{{(-1)}^n-1\}\frac{\pi }{4},n=0,\pm 1,\pm 2$

We have, $\sin \theta +\cos \theta =1$

Multiplying both sides by $\frac{1}{\sqrt{2}}$ , we get

$\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{1}{\sqrt{2}}$

$⟹\sin \theta \cos \frac{\pi }{4}+\cos \theta \sin \frac{\pi }{4}=\sin \frac{\pi }{4}$

$⟹\sin (\theta +\frac{\pi }{4})=\sin \frac{\pi }{4}$

General solution:

$⟹\theta +\frac{\pi }{4}=n\pi +(-1{)}^n\frac{\pi }{4}$

$⟹\theta =n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{4}$

$⟹\theta =n\pi +\left\{\right(-1{)}^n-1\}\frac{\pi }{4},n=0,\pm 1,\pm 2$

Hence, $C$ is the correct option.