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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.

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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