Answer
453.6k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)