The general solution of $\sin x+\cos x=1$ is
The general solution of $\sin x+\cos x=1$ is
- $x=2 \mathrm{n} \pi, \mathrm{n} \in \mathbb{Z}$
- $\quad x=2 \mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathbb{Z}$
- $\quad x=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}-\frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}$
- not existing
Solution
$\begin{aligned}
& \sin x+\cos x=1 \\
& \sqrt{2}\left[\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}}\right]=1
\end{aligned}$
$\begin{aligned}
& \sin x \cdot \cos 45^{\circ}+\cos x \cdot \sin 45^{\circ}=\frac{1}{\sqrt{2}} \\
& \sin \left(x+\frac{\pi}{4}\right)=\sin \left(n \pi+(-1)^n \frac{\pi}{4}\right)
\end{aligned}$
Hence
$\begin{aligned}
& \mathrm{x}+\frac{\pi}{4}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4} \\
& \mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}-\frac{\pi}{4}
\end{aligned}$
Asked in: MHT CET 2024 (11 May Shift 2)
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