Question

The general solution of sinx – cosx = $\sqrt 2 $, for any integer n is
A. ${\text{n}}\pi $
B. $2{\text{n}}\pi + \dfrac{{3\pi }}{4}$
C. $2{\text{n}}\pi $
D. $(2{\text{n - 1)}}\pi $

Answer 1

Hint: First we divide both sides by $\sqrt {{1^2} + {1^2}} {\text{ i}}{\text{.e}}{\text{. }}\sqrt 2 $. Then we write $\dfrac{1}{{\sqrt 2 }}$in terms of ${\text{sin4}}{{\text{5}}^0}$ and ${\text{cos4}} {{\text{5}}^0}$. After this we will simplify the LHS using the formula of cos(A+B). At last we will use the formula for finding the general solution of cosx = ${\text{cos}}\theta $.

Complete Step-by-Step solution:
The given equation is:
sinx – cosx = $\sqrt 2 $. --------------- (1)
Comparing with asinx + bcosx = c, we get:
a = 1
b = -1 and
c = $\sqrt 2 $.
Dividing both sides of equation 1 by $\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} {\text{ i}}{\text{.e }}\sqrt {{1^2} + {1^2}} = \sqrt 2 $, we get:
$\dfrac{1}{{\sqrt 2 }}{\text{sinx - }}\dfrac{1}{{\sqrt 2 }}{\text{cosx = }}\dfrac{{\sqrt 2 }}{{\sqrt 2 }} = 1$ ------------------- (2)
We know that ${\text{sin13}}{{\text{5}}^0} = \dfrac{1}{{\sqrt 2 }}{\text{ and cos13}}{{\text{5}}^0} = - \dfrac{1}{{\sqrt 2 }}$ and also ${\text{cos}}{{\text{0}}^0} = 1$.
So putting these values in equation 2, we get:
${\text{sin13}}{{\text{5}}^0}{\text{sinx + cos13}}{{\text{5}}^0}{\text{cosx = cos}}{{\text{0}}^0}$
$ \Rightarrow \cos {135^0}{\text{cosx + sin13}}{{\text{5}}^0}{\text{sinx }} = {\text{cos}}{{\text{0}}^0}$ . -------------- (3)

We know that Cos(A-B) = CosACosB + SinASinB.
Therefore, the above equation can be written as:
${\text{cos}}\left( {x - {{135}^0}} \right) = {\text{cos}}{{\text{0}}^0}$.
 ${\text{cos}}\left( {x - \dfrac{{3\pi }}{4}} \right) = {\text{cos}}{{\text{0}}^0}$ ------ (4)
Now, the general solution of cosx = ${\text{cos}}\theta $ is given as:
x = $2{\text{n}}\pi \pm \theta {\text{, where }}\theta \in {\text{(0,}}\pi {\text{)}}$
Therefore, the solution of equation 4 can be written as:
$x - \dfrac{{3\pi }}{4} = 2{\text{n}}\pi \pm {\text{0 = 2n}}\pi $
$ \Rightarrow x = {\text{2n}}\pi + \dfrac{{3\pi }}{4}$ .
Hence, option B is correct.

Note: In the question involving the solution of trigonometric equation, you have to convert the given equation in some standard for whose general is known. Some of the general solution that you should remember are:
$
  {\text{1}}{\text{. sinx = 0 }} \Rightarrow {\text{ x = n}}\pi \\
  2.\cos x = 0 \Rightarrow {\text{ x = n}}\pi + \pi /2 \\
  3.\tan x = 0 \Rightarrow {\text{ x = n}}\pi \\
  4.\cos x = \cos \theta \Rightarrow {\text{ x = 2n}}\pi \pm \theta ,{\text{where }}\theta \in {\text{(0,}}\pi {\text{)}} \\
  {\text{5}}{\text{.sin}}x = \sin \theta \Rightarrow {\text{ x = n}}\pi \pm {( - 1)^n}\theta ,{\text{where }}\theta \in {\text{( - }}\pi {\text{/2,}}\pi /2{\text{)}} \\
  {\text{6}}{\text{.tan}}x = \tan \theta \Rightarrow {\text{ x = n}}\pi + \theta ,{\text{where }}\theta \in {\text{( - }}\pi {\text{/2,}}\pi /2{\text{)}} \\
 $