Question

The general solution of the trigonometric equation sin x+ cos x =1 is given by

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

Answer 1

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

sin x + cos x =1
or $\frac{1}{\sqrt{2}}sinx+\frac{1}{\sqrt{2}}cosx=\frac{1}{\sqrt{2}}$
or $sinxcos\left(\frac{\pi }{4}\right)+cosxsin\left(\frac{\pi }{4}\right)=sin\frac{\pi }{4}$
or $sin(x+\frac{\pi }{4})=sin\frac{\pi }{4}$
$\Rightarrow x+\left(\frac{\pi }{4}\right)=n\pi +(-1{)}^n\frac{\pi }{4},n\in Z$
$\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{4}$
where $n=0,\pm 1,\pm 2,.....$