Question

How do you solve $\sin x+\cos x=1$?

Answer 1

Find the general solution of given trigonometric equation

Given: $\sin x+\cos x=1$

Squaring on both sides we get,

$\Rightarrow$ $\mathrm{sinx}+\mathrm{cosx}=1$

$\Rightarrow$ $\frac{1}{\sqrt{2}}\mathrm{sinx}+\frac{1}{\sqrt{2}}\cos x=\frac{1}{\sqrt{2}}$

$\Rightarrow$ $\sin \frac{\mathrm{\pi }}{4}\sin x+\cos x\cos \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}$

$\Rightarrow$ $\cos (x-\frac{\mathrm{\pi }}{4})=\frac{1}{\sqrt{2}}$

$\Rightarrow$ $x-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}$

$\Rightarrow$ $x-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\Rightarrow \mathrm{x}=2\mathrm{n\pi }(\mathrm{taking}+\mathrm{sign})$

$\Rightarrow$ $x-\frac{\mathrm{\pi }}{4}=2n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\Rightarrow \mathrm{x}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2}(\mathrm{taking}-\mathrm{sign})$

$\Rightarrow$ $x=\frac{\mathrm{\pi }}{2}+2\mathrm{\pi n},2\mathrm{\pi n}$ { $n$ is any integer}

Hence the answer is $x=\frac{\mathrm{\pi }}{2}+2\mathrm{\pi n},2\mathrm{\pi n}$ { $n$ is any integer}