Question

The general solution for the equation $\cos x+\sin x=\sqrt{2}$ is ____.

Answer 1

Given equation: $\cos x+\sin x=\sqrt{2}$
$\Rightarrow \sin x+\cos x=\sqrt{2}$

Comparing with the standard equation $a\sin x+b\cos x=c,$

$a=1\ne 0;b=1\ne 0;c=\sqrt{2};$
$\frac{c}{\sqrt{a^2+b^2}}=\frac{\sqrt{2}}{\sqrt{1^2+1^2}}=\sqrt{\frac{2}{2}}=1$

$\therefore -1\le \frac{c}{\sqrt{a^2+b^2}}\le 1,$ which signifies the given eqaution has a valid solution.

Divide the equation by $\sqrt{a^2+b^2}=\sqrt{2}.$
$\Rightarrow \frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x=\frac{\sqrt{2}}{\sqrt{2}}$
Here, the auxiliary angle is $\frac{\pi }{4}.$

Substitute $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}},\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
$\Rightarrow \cos \frac{\pi }{4}\sin x+\sin \frac{\pi }{4}\cos x=1$
Apply $\sin (A+B)=\sin A\cos B+\cos A\sin B$

$\Rightarrow \sin (x+\frac{\pi }{4})=1=\sin \frac{\pi }{2}$

The general solution is $x+\frac{\pi }{4}=n\pi +(-1{)}^n\frac{\pi }{2},n\in Z$
$\Rightarrow x=n\pi -\frac{\pi }{4}+(-1{)}^n\frac{\pi }{2},n\in Z$
$\Rightarrow x=(4n-1)\frac{\pi }{4}+(-1{)}^n\frac{\pi }{2},n\in Z$