Question

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

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

Answer 1

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

$sinx+cosx=1$
$\sqrt{2}[\frac{sinx}{\sqrt{2}}+\frac{cosx}{\sqrt{2}}]=1$

$sinx.cos{45}^0+cosx.sin{45}^0=\frac{1}{\sqrt{2}}$

$sin(x+\frac{\pi }{4})=sin(n\pi +(-1{)}^n\frac{\pi }{4})$

Hence

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

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