Question

The general solution of \[\sin x-\cos x=\sqrt{2}\] for any integer \[n\] is
(A) \[n\pi \]
(B) \[2n\pi +\dfrac{3\pi }{4}\]
(C) \[2n\pi \]
(D) \[\left( 2n+1 \right)\pi \]

Answer 1

Hint: Divide the LHS and RHS of the expression \[\sin x-\cos x=\sqrt{2}\] by \[\sqrt{2}\] . We know that \[\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] . Now, use the formula, \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\] and simplify the given expression. We also know that \[\sin \dfrac{\pi }{2}=1\] and \[\sin \dfrac{5\pi }{2}=1\] . Now, use the identity \[\sin \left( {{\sin }^{-1}}\theta \right)=\theta \] and get the values of \[x\] . At last, generalize the value of \[x\] .

Complete step by step answer:
According to the question, we are given a trigonometric expression and we have to find its general solution.
The given expression is \[\sin x-\cos x=\sqrt{2}\] ………………………………….(1)
Now, on dividing the LHS and RHS of equation (1) by \[\sqrt{2}\] , we get
\[\Rightarrow \sin x.\dfrac{1}{\sqrt{2}}-\cos x.\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{\sqrt{2}}\]
\[\Rightarrow \left( \sin x.\dfrac{1}{\sqrt{2}}-\cos x.\dfrac{1}{\sqrt{2}} \right)=1\] ……………………………………..(2)
We know that \[\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] …………………………………………..(3)
Now, from equation (2) and equation (3), we get
\[\Rightarrow \left( \sin x.\cos \dfrac{\pi }{4}-\cos x.\sin \dfrac{\pi }{4} \right)=1\] ………………………………………………..(4)
We know the formula, \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\] ………………………………………………(5)
Now, from equation (4) and equation (5), we get
\[\Rightarrow \sin \left( x-\dfrac{\pi }{4} \right)=1\] …………………………………….(6)
We also know that \[\sin \dfrac{\pi }{2}=1\] ……………………………………(7)
Now, using equation (7) and on simplifying equation (6), we get
 \[\Rightarrow \sin \left( x-\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{2}\]
\[\Rightarrow \left( x-\dfrac{\pi }{4} \right)={{\sin }^{-1}}\left( \sin \dfrac{\pi }{2} \right)\] ……………………………..(8)
We know the identity, \[\sin \left( {{\sin }^{-1}}\theta \right)=\theta \] ………………………………………..(9)
Now, from equation (8) and equation (9), we get
\[\begin{align}
  & \Rightarrow \left( x-\dfrac{\pi }{4} \right)=\dfrac{\pi }{2} \\
 & \Rightarrow x=\dfrac{\pi }{4}+\dfrac{\pi }{2} \\
\end{align}\]
\[\Rightarrow x=\dfrac{3\pi }{4}\] ………………………………………..(10)
We also know that \[\sin \dfrac{5\pi }{2}=1\] ……………………………………(11)
Now, using equation (11) and on simplifying equation (6), we get
 \[\Rightarrow \sin \left( x-\dfrac{\pi }{4} \right)=\sin \dfrac{5\pi }{2}\]
\[\Rightarrow \left( x-\dfrac{\pi }{4} \right)={{\sin }^{-1}}\left( \sin \dfrac{5\pi }{2} \right)\] ……………………………..(12)
From equation (9) and equation (12), we get
\[\begin{align}
  & \Rightarrow \left( x-\dfrac{\pi }{4} \right)=\dfrac{5\pi }{2} \\
 & \Rightarrow x=\dfrac{\pi }{4}+\dfrac{\pi }{2}+2\pi \\
\end{align}\]
\[\Rightarrow x=2\pi +\dfrac{3\pi }{4}\] ………………………………………………(13)
Now, from equation (10) and equation (13), we have the value of \[x\] i.e., \[x=\dfrac{3\pi }{4}\] and \[x=2\pi +\dfrac{3\pi }{4}\] …………………………………(14)
On generalizing equation (14), we can say that
\[x=2n\pi +\dfrac{3\pi }{4}\] ……………………………….(15)
Therefore, the general value of \[\sin x-\cos x=\sqrt{2}\] is \[x=2n\pi +\dfrac{3\pi }{4}\] .
Hence, the correct option is (B).

Note:
 Whenever this question appears where we are given a sine and cosine expression in LHS and some numerical value in RHS and we are asked the general value. Then, always this type of question by making the numerical value in RHS less than or equal to 1 by using some mathematical operations.