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If $\sin \theta+\cos \theta=1$, then find the general value of $\theta$.
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Verified Answer
Given that, $\sin \theta+\cos \theta=1$
$$
\Rightarrow \frac{1}{\sqrt{2}} \cdot \sin \theta+\frac{1}{\sqrt{2}} \cdot \cos \theta=\frac{1}{\sqrt{2}}
$$
$$
\begin{aligned}
&\Rightarrow \sin \theta \cdot \cos \frac{\pi}{4}+\cos \theta \cdot \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\
&\Rightarrow \sin \left(\theta+\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \\
&\qquad \because \sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y] \\
&\Rightarrow \theta+\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{4} \\
&\Rightarrow \theta=n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4}
\end{aligned}
$$
$$
\Rightarrow \frac{1}{\sqrt{2}} \cdot \sin \theta+\frac{1}{\sqrt{2}} \cdot \cos \theta=\frac{1}{\sqrt{2}}
$$
$$
\begin{aligned}
&\Rightarrow \sin \theta \cdot \cos \frac{\pi}{4}+\cos \theta \cdot \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\
&\Rightarrow \sin \left(\theta+\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \\
&\qquad \because \sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y] \\
&\Rightarrow \theta+\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{4} \\
&\Rightarrow \theta=n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4}
\end{aligned}
$$
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