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The general solution of the trigonometric equation \[\sin x+\cos x=1\], for \[n=0,\pm 1,\pm 2,\pm 3..........\] is given by:
(a) \[x=2n\pi \]
(b) \[x=2n\pi +\dfrac{1}{2}\pi \]
(c) \[x=n\pi +{{(-1)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{4}\]
(d) None of these

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Hint: Firstly, consider an expression of the form \[r\sin \left( x+a \right)=1\]. Expand this expression using the formula, \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\] and compare with the given trigonometric function in the question. Upon using the required trigonometric identity of sine function and the general solution, we can compute the answer.

Given, \[\sin x+\cos x=1\], which is a trigonometric equation.
This is a problem based on finding out the general solution of a trigonometric equation.
To initiate the process, let us assume a trigonometric equation:
\[r\sin \left( x+a \right)=1\].
\[r\sin \cos a+r\cos x\sin a=1\].
We have written the above equation using the trigonometry compound angle formula which is given below:
\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.\]
Now let us compare both the equations:
\[r\sin \cos a+r\cos x\sin a=1\]with \[\sin x+\cos x=1\].
Through that we have:
\[\begin{align}
  & r\cos a=1 \\
 & \Rightarrow \cos a=\dfrac{1}{r}.........(i) \\
 & r\sin a=1 \\
 & \Rightarrow \sin a=\dfrac{1}{r}.........(ii) \\
\end{align}\]
Dividing the above-mentioned equations, we get:
\[\dfrac{\sin a}{\cos a}=\dfrac{\dfrac{1}{r}}{\dfrac{1}{r}}\]
\[\Rightarrow tana=1\]
But we know, $\tan \dfrac{\pi }{4}=1$
So, the value of a is \[\dfrac{\pi }{4}\].
As we know that \[{{\cos }^{2}}a+{{\sin }^{2}}a=1\], substituting values from equation (i) and (ii), we can rewrite this equation as:
\[\dfrac{1}{{{r}^{2}}}+\dfrac{1}{{{r}^{2}}}=1\].
\[\Rightarrow \dfrac{2}{{{r}^{2}}}=1\]
\[{{r}^{2}}=2\]
Taking the square root on both sides, we get
\[r=\sqrt{2}\].
Substituting the value of ‘r’ and ‘a’ in the assumed equation\[r\sin \left( x+a \right)=1\], we have:
\[\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)=1\]
\[\sin \left( x+\dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\]
But we know, $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ , substituting this value the above equation can be written as,
\[\sin \left( x+\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{4}\]
The general solution of this equation is
\[\Rightarrow x+\dfrac{\pi }{4}=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}\], for \[n=0,\pm 1,\pm 2,\pm 3..........\]
Therefore \[x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{4}\] where \[n\in z\], because if \[\sin x=\sin y\] then x is given as \[x=n\pi +{{\left( -1 \right)}^{n}}y\].
Hence, the correct answer is option (c).

Note: Alternatively, you can directly multiply and divide the given trigonometric equation simultaneously with \[\sqrt{2}\] and achieve \[\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)=1\] directly, thus saving time. Using identities in solving trigonometric functions plays a key role in these types of questions.
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