Question

Solve $\sin x+\cos x=\sqrt{2}$, then $x=2n\pi +\frac{\pi }{4},n\in I$
If true then enter $1$ and if false then enter $0$

Answer 1

$\because \sin x+\cos x=\sqrt{2}$
Here a = 1, b = 1.
$\therefore$ divide both sides of equation (i) by $\sqrt{2},$ we get
$\sin x\frac{1}{\sqrt{2}}+\cos x\frac{1}{\sqrt{2}}=1$

$\Rightarrow \sin x\sin \frac{\pi }{4}+\cos x.\cos \frac{\pi }{4}=1$

$\Rightarrow \cos (x-\frac{\pi }{4})=1$

$\Rightarrow x-\frac{\pi }{4}=2n\pi ,n\in I$

$\Rightarrow x=2n\pi +\frac{\pi }{4},n\in I$

$\therefore$ Solution of given equation is $2n\pi +\frac{\pi }{4},n\in I$