# 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

Last updated date: 25th Mar 2023

•

Total views: 308.7k

•

Views today: 2.85k

Answer

Verified

308.7k+ 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

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE